CHAPTER
BERNOULLI AND ENERGY
E Q U AT I O N S
his chapter deals with two equations commonly used in fluid mechanics: the Bernoulli equation and the energy equation. The Bernoulli equation is concerned with the conservation of kinetic, potential, and flow
energies of a fluid stream, and their conversion to each other in regions of
flow where net viscous forces are negligible, and where other restrictive conditions apply. The energy equation is a statement of the conservation of energy principle and is applicable under all conditions. In fluid mechanics, it is
found to be convenient to separate mechanical energy from thermal energy
and to consider the conversion of mechanical energy to thermal energy as a result of frictional effects as mechanical energy loss. Then the energy equation
is usually expressed as the conservation of mechanical energy.
We start this chapter with a discussion of various forms of mechanical energy and the efficiency of mechanical work devices such as pumps and turbines. Then we derive the Bernoulli equation by applying Newton’s second
law to a fluid element along a streamline and demonstrate its use in a variety
of applications. We continue with the development of the energy equation in
a form suitable for use in fluid mechanics and introduce the concept of head
loss. Finally, we apply the energy equation to various engineering systems.
T
12
CONTENTS
12–1 Mechanical Energy and
Efficiency 520
12–2 The Bernoulli Equation 525
12–3 Applications of the Bernoulli
Equation 534
12–4 Energy Analysis of Steady-Flow
Systems 541
Summary 549
References and Suggested
Reading 551
Problems 551
519
520
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
12–1
z
Atmosphere
·
W
h = 10 m
1
0
m· = 2 kg/s
ρgh ·
P1 – Patm
·
= m· ––– = mgh
Wmax = m· –––––––––
ρ
ρ
= (2 kg/s)(9.81 m/s2)(10 m)
= 196 W
FIGURE 12–1
In the absence of any changes in flow
velocity and elevation, the power
produced by a hydraulic turbine is
proportional to the pressure drop of
water across the turbine.
■
MECHANICAL ENERGY AND EFFICIENCY
Most fluid systems are designed to transport a fluid from one location to
another at a specified flow rate, velocity, and elevation difference, and the
system may generate mechanical work in a turbine or it may consume mechanical work in a pump or fan during this process. These systems do not
involve the conversion of nuclear, chemical, or thermal energy to mechanical
energy. Also, they do not involve any heat transfer in any significant amount,
and they operate essentially at constant temperature. Such systems can be analyzed conveniently by considering the mechanical forms of energy only
and the frictional effects that cause the mechanical energy to be lost (i.e., to
be converted to thermal energy that usually cannot be used for any useful
purpose).
The mechanical energy can be defined as the form of energy that can be
converted to mechanical work completely and directly by a mechanical device
such as an ideal turbine. Kinetic and potential energies are the familiar forms
of mechanical energy. Thermal energy is not mechanical energy, however,
since it cannot be converted to work directly and completely (the second law
of thermodynamics).
A pump transfers mechanical energy to a fluid by raising its pressure, and a
turbine extracts mechanical energy from a fluid by dropping its pressure.
Therefore, the pressure of a flowing fluid is also associated with its mechanical energy. In fact, the pressure unit Pa is equivalent to Pa N/m2 N m/m3
J/m3, and the product Pυ or its equivalent P/r has the unit J/kg, which is energy unit per unit mass. Note that pressure itself is not a form of energy. But a
pressure force acting on a fluid through a distance produces work, called flow
work, in the amount of P/r per unit mass. Flow work is expressed in terms of
fluid properties, and it is found convenient to view it as part of the energy of
a flowing fluid, and call it flow energy. Therefore, the mechanical energy of a
flowing fluid can be expressed on a unit mass basis as (Fig. 12–1).
P 2
emech r gz
2
where P/r is the flow energy, 2 /2 is the kinetic energy, and gz is the potential energy of the fluid per unit mass. Then the mechanical energy change of a
fluid during incompressible flow becomes
emech P2 P1 22 21
g(z2 z1)
r 2
(kJ/kg)
(12–1)
Therefore, the mechanical energy of a fluid does not change during flow if its
pressure, density, velocity, and elevation remain constant. In the absence of
any losses, the mechanical energy change represents the mechanical work
supplied to the fluid (if ∆emech 0) or extracted from the fluid (if ∆emech 0).
Consider a container of height h filled with water, as shown in Fig. 12–2,
with reference level selected at the bottom surface. The gage pressure and the
potential energy per unit mass are PA 0 and peA gh at point A at the free
surface, and PB rgh and peB 0 at point B at the bottom of the container.
An ideal hydraulic turbine would produce the same work per unit mass
wturbine gh whether it receives water (or any other fluid with constant
density) from the top or from the bottom of the container. Note that we are
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CHAPTER 12
z
h
P=0
A pe = gh
m·
·
·
Wmax = mgh
0
B P = ρgh
pe = 0
·
·
Wmax = mgh
m·
FIGURE 12–2
The mechanical energy of water at
the bottom of a lake is equal to the
mechanical energy at any depth
including the free surface of the lake.
also assuming ideal flow (no irreversible losses) through the pipe leading from
the tank to the turbine. Therefore, the total mechanical energy of water at the
bottom is equivalent to that at the top.
The transfer of mechanical energy is usually accomplished by a rotating
shaft, and thus mechanical work is often referred to as shaft work. A pump or
a fan receives shaft work (usually from an electric motor) and transfers it to
the fluid as mechanical energy (less frictional losses). A turbine, on the other
hand, converts the mechanical energy of a fluid to shaft work. In the absence
of any irreversibilities such as friction, mechanical energy can be converted
entirely from one mechanical form to another, and the mechanical efficiency
of a device or process can be defined as (Fig. 12–3).
E
E
hmech Mechanical energy output mech, out 1 mech, loss
Emech, in
Emech, in
Mechanical energy input
1
(12–3)
where Wpump, u E mech, fluid E mech, out E mech, in is the rate of increase in the
mechanical energy of the fluid, which is equivalent to the useful pumping
power supplied to the fluid, and
hturbine 50 W
(12–2)
A conversion efficiency of less than 100 percent indicates that conversion is
less than perfect and some losses have occurred during conversion. A mechanical efficiency of 97 percent indicates that 3 percent of the mechanical energy input is converted to thermal energy as a result of frictional heating, and
this will manifest itself as a slight rise in the temperature of the fluid.
In fluid systems, we are usually interested in increasing the pressure, velocity, and/or elevation of a fluid. This is done by supplying mechanical energy
to the fluid by a pump, a fan, or a compressor (we will refer to all of them as
pumps). Or we are interested in the reverse process of extracting mechanical
energy from a fluid by a turbine, and producing mechanical power in the form
of a rotating shaft that can drive a generator or any other rotary device. The
degree of perfection of the conversion process between the mechanical work
supplied or extracted and the mechanical energy of the fluid is expressed by
the pump efficiency and turbine efficiency, defined as
Mechanical energy increase of the fluid W pump, u
hpump Mechanical energy input
Wpump
Fan
Wshaft, out
Mechanical engery output
Wturbine
(12–4)
Mechanical energy decrease of the fluid E mech, fluid Wturbine, e
m· = 0.25 kg/s
2
1 = 0, 2 = 12 m/s
z1 = z2
P1 = P2
·
2
m·2/2
∆Emech, fluid
ηmech, fan = ––––––––––
= –––––––
·
·
Wshaft, in
Wshaft, in
(0.25 kg/s)(12 m/s)2/2
= –––––––––––––––––
50 W
= 0.72
FIGURE 12–3
The mechanical efficiency of a
fan is the ratio of the kinetic energy
of air at the fan exit to the
mechanical power input.
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FUNDAMENTALS OF THERMAL-FLUID SCIENCES
where E mech, fluid E mech, in E mech, out is the rate of decrease in the mechanical energy of the fluid, which is equivalent to the mechanical power ex
tracted from the fluid by the turbine Wturbine, e , and we used the absolute value
sign to avoid negative values for efficiencies. A pump or turbine efficiency of
100 percent indicates perfect conversion between the shaft work and the mechanical energy of the fluid, and this value can be approached (but never attained) as the frictional effects are minimized.
The mechanical efficiency should not be confused with the motor efficiency and the generator efficiency, which are defined as
hmotor Motor:
Mechanical power output Wshaft, out
Electrical power input
Welect, in
(12–5)
and
hgenerator Generator:
ηturbine = 0.75
Turbine
ηgenerator = 0.97
FIGURE 12–4
The overall efficiency of a turbinegenerator is the product of the
efficiency of the turbine and the
efficiency of the generator, and
represents the fraction of the
mechanical energy of the fluid
converted to electrical energy.
(12–6)
A pump is usually packaged together with its motor, and a hydraulic turbine
with its generator. Therefore, we are usually interested in the combined or
overall efficiency of pump/motor and turbine/generator combinations
(Fig. 12–4), which are defined as
W pump, u E mech, fluid
hpump-motor hpump hmotor Welect, in
Welect, in
Generator
ηturbine-gen = ηturbine ηgenerator
= 0.75 × 0.97
= 0.73
Welect, out
Electrical power output
Mechanical power input Wshaft, in
(12–7)
and
Welect, out
hturbine-gen hturbine hgenerator W turbine, e
Welect, out
Emech, fluid
(12–8)
All the efficiencies just defined range between 0 and 100 percent. The lower
limit of 0 percent corresponds to the conversion of the entire mechanical or
electrical energy input to thermal energy, and the device in this case functions
like a resistance heater. The upper limit of 100 percent corresponds to the case
of perfect conversion with no friction or other irreversibilities, and thus no
conversion of mechanical or electrical energy to thermal energy.
EXAMPLE 12–1
Performance of a Hydraulic Turbine-Generator
The water in a large lake is to be used to generate electricity by installing a
hydraulic turbine-generator at a location where the depth of the water is 50 m
(Fig. 12–5). Water is to be supplied at a rate of 5000 kg/s. If the electric power
generated is measured to be 1862 kW and the generator efficiency is 95 percent, determine (a) the overall efficiency of the turbine-generator, (b) the mechanical efficiency of the turbine, and (c) the shaft power supplied by the
turbine to the generator.
SOLUTION
A hydraulic turbine-generator is to generate electricity from the
water of a lake. The overall efficiency, the turbine efficiency, and the shaft
power are to be determined.
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CHAPTER 12
ηgenerator = 0.95
1862 kW
Lake
h = 50 m
Turbine
Generator
m· = 5000 kg/s
Assumptions 1 The elevation of the lake remains constant. 2 The mechanical
energy of water at the turbine exit is negligible.
Properties The density of water can be taken to be r 1000 kg/m3.
Analysis (a) We take the bottom of the lake as the reference level for convenience. Then kinetic and potential energies of water are zero, and the change in
its mechanical energy per unit mass becomes
1 kJ/kg
P
emech, in emech, out r 0 gh (9.81 m/s2)(50 m)a
b 0.491 kJ/kg
1000 m2/s2
Then the rate at which mechanical energy is supplied to the turbine by the fluid
and the overall efficiency become
(e
Emech, fluid m
mech, in emech, out) (5000 kg/s)(0.491 kJ/kg) 2455 kW
Welect, out
1862 kW
0.76
hoverall hturbine-gen Emech, fluid 2455 kW
(b) Knowing the overall and generator efficiencies, the mechanical efficiency of
the turbine is determined from
hturbine-gen hturbine hgenerator
→
hturbinegen 0.76
hturbine h
0.80
generator
0.95
(c) The shaft power output is determined from the definition of mechanical
efficiency,
W shaft, out hturbine Emech, fluid (0.80)(2455 kW) 1964 kW
Discussion Note that the lake supplies 2455 kW of mechanical energy to the
turbine, which converts 1964 kW of it to shaft work that drives the generator,
which generates 1862 kW of electric power. There are irreversible losses
through each component.
EXAMPLE 12–2
Conservation of Energy for an Oscillating
Steel Ball
The motion of a steel ball in a hemispherical bowl of radius h shown in Fig.
12–6 is to be analyzed. The ball is initially held at the highest location at point
A, and then it is released. Obtain relations for the conservation of energy of the
ball for the cases of frictionless and actual motions.
SOLUTION A steel ball is released in a bowl. Relations for the energy balance
are to be obtained.
FIGURE 12–5
Schematic for Example 12–1.
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FUNDAMENTALS OF THERMAL-FLUID SCIENCES
z
A
h
Steel
ball
C
1
FIGURE 12–6
Schematic for Example 12–2.
0
2
B
Assumptions The motion is frictionless, and thus friction between the ball, the
bowl, and the air is negligible.
Analysis When the ball is released, it accelerates under the influence of gravity, reach a maximum velocity (and minimum elevation) at point B at the bottom of the bowl, and move up toward point C on the opposite side. In the ideal
case of frictionless motion, the ball will oscillate between points A and C. The
actual motion involves the conversion of the kinetic and potential energies of
the ball to each other, together with overcoming resistance to motion due to
friction (doing frictional work). The general energy balance for any system undergoing any process is
Ein Eout
1424
3
Net energy transfer
by heat, work, and mass
E system
14243
Change in internal, kinetic,
potential, etc., energies
Then the energy balance for the ball for a process from point 1 to point 2
becomes
wfriction (ke2 pe2) (ke1 pe1)
or
21
22
gz1 gz2 wfriction
2
2
since there is no energy transfer by heat or mass, and no change in the internal
energy of the ball (the heat generated by frictional heating is dissipated to the
surrounding air). The frictional work term wfriction is often expressed as eloss to
represent the loss (conversion) of mechanical energy into thermal energy.
For the idealized case of frictionless motion, the last relation reduces to
21
22
gz1 gz2
2
2
or
2
gz C constant
2
where the value of the constant is C gh. That is, when the frictional effects
are negligible, the sum of the kinetic and potential energies of the ball remains
constant.
Discussion This is certainly a more intuitive and convenient form of the conservation of energy equation for this and other similar processes such as the
swinging motion of the pendulum of a wall clock. The relation obtained is analogous to the Bernoulli equation derived in the next section.
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CHAPTER 12
Most processes encountered in practice involve only certain forms of energy,
and in such cases it is more convenient to work with the simplified versions of
the energy balance. For systems that involve only mechanical forms of energy
and its transfer as shaft work, the conservation of energy principle can be expressed conveniently as
Emech, in Emech, out Emech, system Emech, loss
2
h
(12–9)
·
Wpump
where Emech, loss represents the conversion of mechanical energy to thermal energy due to irreversibilities such as friction. For a system in steady operation, the
mechanical energy balance becomes Emech, in Emech, out Emech, loss (Fig. 12–7).
12–2
■
THE BERNOULLI EQUATION
The Bernoulli equation is an approximate relation between pressure, velocity, and elevation, and is valid in regions of steady, incompressible flow where
net frictional forces are negligible (Fig. 12–8). Despite its simplicity, it has
proven to be a very powerful tool in fluid mechanics. In this section, we derive the Bernoulli equation by applying the conservation of linear momentum
principle, and we demonstrate both its usefulness and its limitations.
The key approximation in the derivation of the Bernoulli equation is that
viscous effects are negligibly small compared to inertial, gravitational, and
pressure effects. Since all fluids have viscosity (there is no such thing as an
“inviscid fluid”), this approximation cannot be valid for an entire flow field of
practical interest. In other words, we cannot apply the Bernoulli equation
everywhere in a flow, no matter how small the fluid’s viscosity. However, it
turns out that the approximation is reasonable in certain regions of many practical flows. We refer to such regions as inviscid regions of flow, and we stress
that they are not regions where the fluid itself is inviscid or frictionless, but
rather they are regions where net viscous or frictional forces are negligibly
small compared to other forces acting on fluid particles.
Care must be exercised when applying the Bernoulli equation since it is an
approximation that applies only to inviscid regions of flow. In general, frictional effects are always important very close to solid walls (boundary layers)
and directly downstream of bodies (wakes). Thus, the Bernoulli approximation is typically useful in flow regions outside of boundary layers and wakes,
where the fluid motion is governed by the combined effects of pressure and
gravity forces.
The motion of a particle and the path it follows are described by the velocity vector as a function of time and space coordinates and the initial position
of the particle. When the flow is steady (no change with time at a specified location), all particles that pass through the same point follow the same path
(which is the streamline), and the velocity vectors remain tangent to the path
at every point.
Acceleration of a Fluid Particle
Often it is convenient to describe the motion of a particle in terms of its distance s from the origin together with the radius of curvature along the streamline. The velocity of the particle is related to the distance by ds/dt, which
may vary along the streamline. In two-dimensional flow, the acceleration can
be decomposed into two components: streamwise acceleration as along the
1
Steady flow
1 = 2
z2 = z1 + h
P1 ≅ P2 ≅ Patm
·
·
·
Emech, in = Emech, out + Emech, loss
·
·
·
·
Wpump + mgz1 = mgz2 + Emech, loss
·
· + E·
Wpump = mgh
mech, loss
FIGURE 12–7
Most fluid flow problems involve
mechanical forms of energy only, and
such problems are conveniently solved
by using a mechanical energy balance.
Bernoulli equation valid
Bernoulli equation not valid
FIGURE 12–8
The Bernoulli equation is an
approximate equation that is valid
only in inviscid regions of flow where
net viscous forces are negligibly small
compared to inertial, gravitational, or
pressure forces. Such regions occur
outside of boundary layers and wakes.
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FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Nozzle
Water
FIGURE 12–9
During steady flow, a fluid may not
accelerate in time at a fixed point,
but it may accelerate in space.
streamline and normal acceleration an in the direction normal to the streamline, which is given as an 2/R. Note that streamwise acceleration is due to
a change in speed along a streamline, and normal acceleration is due to a
change in direction. For particles that move along a straight path, an 0 since
the radius of curvature is infinity and thus there is no change in direction. The
Bernoulli equation results from a force balance along a streamline.
One may be tempted to think that acceleration is zero in steady flow since
acceleration is the rate of change of velocity with time, and in steady flow
there is no change with time. Well, a garden hose nozzle tells us that this understanding is not correct. Even in steady flow and thus constant mass flow
rate, water accelerates through the nozzle (Fig. 12–9). Steady simply means
no change with time at a specified location, but the value of a quantity may
change from one location to another. In the case of a nozzle, the velocity of
water remains constant at a specified point, but it changes from the inlet to the
exit (water accelerates along the nozzle).
Mathematically, this can be expressed as follows: We take the velocity to
be a function of s and t. Taking the total differential of (s, t) and dividing
both sides by dt give
d ds dt
s
t
d ds s dt
t
dt
and
(12–10a, b)
In steady flow /t 0 and thus (s), and the acceleration in the
s direction becomes
as d
d ds s dt
s
dt
ds
(12–11)
Therefore, acceleration in steady flow is due to the change of velocity with
position.
Derivation of the Bernoulli Equation
Consider the motion of a fluid particle in a flow field in steady flow. Applying
Newton’s second law (which is referred to as the conservation of linear momentum relation in fluid mechanics) in the s direction on a particle moving
along a streamline gives
F ma
s
(12–12)
s
In regions of flow where net fictional forces are negligible, the significant
forces acting in the s direction are the pressure (acting on both sides) and the
component of the weight of the particle in the s direction (Fig. 12–10). Therefore, Eq. 12–12 becomes
P dA (P dP) dA W sin u m
d
ds
(12–13)
where u is the angle between the normal of the streamline and the vertical z
axis at that point, m rV r dA ds is the mass, W mg rg dA ds is the
weight of the fluid particle, and sin u dz/ds. Substituting,
dP dA rg dA ds
dz
d
r dA ds ds
ds
(12–14)
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CHAPTER 12
z
Streamline
(P + dP) dA
ds
θ
PdA
ds
n
dz
θ
W
dx
s
x
FIGURE 12–10
The forces acting on a
fluid particle along a streamline.
Canceling dA from each term and simplifying,
dP rg dz r d
(12–15)
Noting that d d(2) and dividing each term by r gives
1
2
(Steady flow)
General:
dP 1
2
r 2 d( ) g dz 0
(12–16)
dP 2
r 2 gz constant (along a streamline)
(12–17)
since the last two terms are exact differentials. In the case of incompressible
flow, the first term also becomes an exact differential, and its integration gives
Steady, incompressible flow:
P 2
r 2 gz constant
(kJ/kg)
(12–18)
This is the famous Bernoulli equation, which is commonly used in fluid mechanics for steady, incompressible flow in inviscid regions of flow. The value
of the constant can be evaluated at any point on the streamline where the pressure, density, velocity, and elevation are known. The Bernoulli equation can
also be written between any two points on the same streamline as
Steady, incompressible flow:
2
Incompressible flow (ρ = constant):
Integrating (Fig. 12–11),
Steady flow:
+ gz = constant
∫ ––dP
ρ + ––
2
P1 21
P2 22
gz
1
r
r 2 gz2
2
(12–19)
The Bernoulli equation is obtained from the conservation of momentum for
a fluid particle moving along a streamline. It can also be obtained from the
first law of thermodynamics applied to a steady-flow system, as shown later in
this chapter.
The Bernoulli equation was first stated verbally in a textbook by Daniel
Bernoulli in 1738 and was derived later by Leonhard Euler in 1755. We recognize 2/2 as kinetic energy, gz as potential energy, and P/r as flow energy per
unit mass. Therefore, the Bernoulli equation can be viewed as an expression of
mechanical energy balance and can be stated as follows (Fig. 12–12):
P 2 + gz = constant
––
ρ + ––
2
FIGURE 12–11
The Bernoulli equation is derived
assuming incompressible flow, and
thus it should not be used for flows
with significant compressibility
effects.
Flow
energy
Potential
energy
P 2 + gz = constant
––
ρ + ––
2
Kinetic
energy
FIGURE 12–12
The Bernoulli equation states that
the sum of the kinetic, potential, and
flow energies of a fluid particle is
constant along a streamline
during steady flow.
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FUNDAMENTALS OF THERMAL-FLUID SCIENCES
The sum of the kinetic, potential, and flow energies of a fluid particle is
constant along a streamline during steady flow when the compressibility
and frictional effects are negligible.
The kinetic, potential, and flow energies are the mechanical forms of
energy, as discussed earlier, and the Bernoulli equation can be viewed as
the “conservation of mechanical energy principle.” This is equivalent to the
general conservation of energy principle for systems that do not involve any
conversion of mechanical energy and thermal energy to each other, and thus
the mechanical energy and thermal energy are conserved separately. The
Bernoulli equation states that during steady, incompressible flow with negligible friction, the various forms of mechanical energy are converted to each
other, but their sum remains constant. In other words, there is no dissipation
of mechanical energy during such flows since there is no friction that converts
mechanical energy to sensible thermal (internal) energy.
Recall that energy is transferred to a system as work when a force is applied
to a system through a distance. In the light of Newton’s second law of motion,
the Bernoulli equation can also be viewed as the work done by the pressure
and gravity forces on the fluid particle is equal to the increase in the kinetic
energy of the particle.
Despite the highly restrictive approximations used in its derivation, the
Bernoulli equation is commonly used in practice since a variety of practical
fluid flow problems can be analyzed to reasonable accuracy with it. This is because many flows of practical engineering interest are steady (or at least
steady in the mean), compressibility effects are relatively small, and net frictional forces are negligible in regions of interest in the flow.
Unsteady Compressible Flow
Similarly, using both terms in the acceleration expression (Eq. 12–10), it can
be shown that the Bernoulli equation for unsteady, compressible flow is
gz constant
ds dPr t
2
2
Unsteady, compressible flow:
(12–20)
Force Balance across Streamlines
A
C
It is left as an exercise to show that a force balance in the direction n normal
to the streamline yields the following relation applicable across the streamlines for steady incompressible flow:
2
P
r R dn gz constant
B
Stationary fluid
D
Flowing fluid
PA = PC
PB = PD
FIGURE 12–13
The variation of pressure with
elevation in steady incompressible
flow along a straight line is the
same as that in the stationary fluid
(but this is not the case for a curved
flow section).
(across streamlines)
(12–21)
For flow along a straight line, R → and thus the relation above reduces
to P/r gz constant or P rgz constant, which is an expression for
the variation of hydrostatic pressure with vertical distance for a stationary
fluid body. Therefore, the variation of pressure with elevation in steady
incompressible flow along a straight line is the same as that in the stationary
fluid (Fig. 12–13).
Static, Dynamic, and Stagnation Pressures
The Bernoulli equation states that the sum of the flow, kinetic, and potential energies of a fluid particle along a streamline is constant. Therefore, the
529
CHAPTER 12
kinetic and potential energies of the fluid can be converted to flow energy
(and vice versa) during flow, causing the pressure to change. This phenomenon can be made more visible by multiplying the Bernoulli equation by the
density r,
Pr
2
rgz constant
2
(kPa)
(12–22)
Each term in this equation has pressure units, and thus each term represents
some kind of pressure:
• P is the static pressure (it does not incorporate any dynamic effects); it
represents the actual pressure of the fluid. This is the same as the pressure
used in thermodynamics and property tables.
• r2/2 is the dynamic pressure; it represents the pressure rise when the
fluid in motion is brought to a stop isentropically.
• rgz is the hydrostatic pressure, which is not pressure in a real sense
since its value depends on the reference level selected; it accounts for the
elevation effects, i.e., of fluid weight on pressure.
The sum of the static, dynamic, and hydrostatic pressures is called the total
pressure. Therefore, the Bernoulli equation states that the total pressure along
a streamline is constant.
The sum of the static and dynamic pressures is called the stagnation pressure, and it is expressed as
Pstag P r
2
2
(kPa)
(12–23)
The stagnation pressure represents the pressure at a point where the fluid is
brought to a complete stop isentropically. The static, dynamic, and stagnation
pressures are shown in Fig. 12–14. When static and stagnation pressures are
measured at a specified location, the fluid velocity at that location can be calculated from
2(Pstag P)
r
B
Piezometer
ρ
––
2
2
(12–24)
Equation 12–24 is useful in the measurement of flow velocity when a combination of a static pressure tap and a Pitot tube is used, as illustrated in Fig.
12–14. A static pressure tap is simply a small hole drilled into a wall such
that the plane of the hole is parallel to the flow direction. It measures the static pressure. A Pitot tube is a small tube with its open end aligned into the
flow so as to sense the full impact pressure of the flowing fluid. It measures
the stagnation pressure. In situations in which the static and stagnation pressure of a flowing liquid are greater than atmospheric pressure, a vertical transparent tube called a piezometer tube (or simply a piezometer) can be
attached to the pressure tap and to the Pitot tube, as sketched. The liquid rises
in the piezometer tube to a column height (head) that is proportional to the
pressure being measured. If the pressures to be measured are below atmospheric, or if measuring pressures in gases, piezometer tubes do not work.
However, the static pressure tap and Pitot tube can still be used, but they must
be connected to some other kind of pressure measurement device such as a
U-tube manometer or a pressure transducer.
Static
pressure, P
Dynamic
pressure
Stagnation
pressure, Pstag
Pitot
tube
Stagnation
point
=
√ 2 (P ρ
stag
– P)
FIGURE 12–14
The static, dynamic, and
stagnation pressures.
530
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
High
Correct
Low
FIGURE 12–15
Careless drilling of the static pressure
tap may result in erroneous reading of
the static pressure.
Stagnation streamline
FIGURE 12–16
Streaklines produced by colored fluid
introduced upstream of an airfoil;
since the flow is steady, the streaklines
are the same as streamlines and
pathlines. The stagnation streamline is
marked. (Courtesy, ONERA).
When the static pressure is measured by drilling a hole in the tube wall, care
must be exercised to ensure that the opening of the hole is flush with the wall
surface, with no extrusions before or after the hole (Fig. 12–15). Otherwise
the reading will incorporate some dynamic effects, and thus it will be in error.
When a stationary body is immersed in a flowing stream, the fluid is
brought to a stop at the nose of the body (the stagnation point). The flow
streamline that extends from far upstream to the stagnation point is called the
stagnation streamline (Fig. 12–16). For a two-dimensional flow in the x-y
plane, the stagnation point is actually a line parallel the z-axis, and the stagnation streamline is actually a surface that separates fluid that flows over the
body from fluid that flows under the body. In an incompressible flow, the fluid
decelerates nearly isentropically from its freestream value to zero at the stagnation point, and the pressure at the stagnation point is thus the stagnation
pressure.
Limitations on the Use of the Bernoulli Equation
The Bernoulli equation is one of the most frequently used and misused equations in fluid mechanics. Its versatility, simplicity, and ease of use make it a
very valuable tool for use in analysis, but the same attributes also make it very
tempting to misuse. Therefore, it is important to understand the restrictions on
its applicability and observe the limitations on its use, as explained below:
1. Steady flow The first limitation on the Bernoulli equation is that it
is applicable to steady flow. Therefore, it should not be used during the
transient start-up and shut-down periods, or during periods of change in
the flow conditions. Note that there is an unsteady form of the Bernoulli
equation, discussion of which is beyond the scope of the present text
(see Panton, 1996).
2. Frictionless flow Every flow involves some friction, no matter how
small, and frictional effects may or may not be negligible. The situation
is complicated even more by the amount of error that can be tolerated.
In general, frictional effects are negligible for short flow sections with
large cross sections, especially at low flow velocities. Frictional effects
are usually significant in long and narrow flow passages, in the wake
region downstream of an object, and in diverging flow sections such as
diffusers because of the increased possibility of the fluid separating from
the walls in such geometries. Frictional effects are also significant near
solid surfaces, and thus the Bernoulli equation is usually applicable
along a streamline in the core region of the flow, but not along a
streamline close to the surface (Fig. 12–17).
A component that disturbs the streamlined structure of flow and thus
causes considerable mixing and back flow such as a sharp entrance of a
tube or a partially closed valve in a flow section can make the Bernoulli
equation inapplicable.
3. No shaft work The Bernoulli equation was derived from a force
balance on a particle moving along a streamline. Therefore, the
Bernoulli equation is not applicable in a flow section that involves a
pump, turbine, fan, or any other machine or impeller since such devices
destroy the streamlines and carry out energy interactions with the fluid
531
CHAPTER 12
Sudden
expansion
1
Long and narrow
tubes
1
2
2
1
2 Flow through
a valve
1
A fan
2
1
2
Boundary layers
Wakes
A heating section
FIGURE 12–17
Frictional effects and
components that disturb the
streamlined structure of flow
in a flow section make the
Bernoulli equation invalid.
particles. When the flow section considered involves any of these
devices, the energy equation should be used instead to account for the
shaft work input or output. However, the Bernoulli equation can still be
applied to a flow section prior to or past a machine (assuming, of course,
that the other restrictions on its use are satisfied). In such cases, the
Bernoulli constant changes from upstream to downstream of the device.
4. Incompressible flow One of the assumptions used in the derivation
of the Bernoulli equation is that r constant and thus the flow is
incompressible. This condition is satisfied by liquids and also by gases
at Mach numbers less than about 0.3 since compressibility effects and
thus density variations of gases are negligible at such relatively low
velocities. Note that there is a compressible form of the Bernoulli
equation (Eq. 12–20).
5. No heat transfer The density of a gas is inversely proportional to
temperature, and thus the Bernoulli equation should not be used for flow
sections that involve significant temperature change such as heating or
cooling sections.
6. Flow along a streamline Strictly speaking, the Bernoulli equation
P/r 2/2 gz C is applicable along a streamline, and the value of
the constant C, in general, is different for different streamlines. But
when a region of the flow is irrotational, and thus there is no vorticity in
the flow field, the value of the constant C remains the same for all
streamlines, and, therefore, the Bernoulli equation becomes applicable
across streamlines as well (Fig. 12–18). Therefore, we do not need to be
concerned about the streamlines when the flow is irrotational, and we
can apply the Bernoulli equation between any two points in the
irrotational region of the flow.
We derived the Bernoulli equation by considering two-dimensional flow
in the x-z plane for simplicity, but the equation is valid for general threedimensional flow as well, as long as it is applied along the same streamline.
1
2
Streamlines
P 12
P 22
––1 + ––
+ gz1 = ––2 + ––
+ gz2
ρ
ρ
2
2
FIGURE 12–18
When the flow is irrotational, the
Bernoulli equation becomes applicable
between any two points along the flow
(not just on the same streamline).
532
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
We should always keep in mind the assumptions used in the derivation of the
Bernoulli equation and make sure that they are not violated.
Hydraulic Grade Line (HGL) and
Energy Grade Line (EGL)
It is often convenient to represent the level of mechanical energy graphically
using heights to facilitate visualization of the various terms of the Bernoulli
equation. This is done by dividing each term of the Bernoulli equation by g
to give
P 2
rg 2g z H constant
(m)
(12–25)
Each term in this equation has the dimension of length and represents some
kind of “head” of a flowing fluid as follows:
• P/rg is the pressure head; it represents the height of a fluid column that
produces the static pressure P.
• 2/2g is the velocity head; it represents the elevation needed for a fluid
to reach the velocity during frictionless free fall.
• z is the elevation head; it represents the potential energy of the fluid.
Pressure
head
Elevation
head
2
P + ––
–– + z = H = constant
ρ
g 2g
Velocity
head
Total head
FIGURE 12–19
An alternative form of the Bernoulli
equation is expressed in terms of
heads as the sum of the pressure,
velocity, and elevation heads is
constant along a streamline.
Also, H is the total head for the flow. Therefore, the Bernoulli equation
can be expressed in terms of heads as: the sum of the pressure, velocity, and
elevation heads along a streamline is constant during steady flow when the
compressibility and frictional effects are negligible (Fig. 12–19).
If a piezometer (measures static pressure) is tapped into a pipe, as shown in
Fig. 12–20, the liquid would rise to a height of P/rg above the pipe center.
The hydraulic grade line (HGL) is obtained by doing this at several locations
along the pipe and drawing a line through the liquid levels in piezometers. The
vertical distance above the pipe center is a measure of pressure within the
pipe. Similarly, if a pitot tube (measures static dynamic pressure) is tapped
into a pipe, the liquid would rise to a height of P/rg 2/2g above the pipe
center, or a distance of 2/2g above the HGL. The energy grade line (EGL) is
obtained by doing this at several locations along the pipe and drawing a line
through the liquid levels in pitot tubes.
Noting that the fluid also has elevation head z (unless the reference level is
taken to be the centerline of the pipe), the HGL and EGL can be defined as
follows: The line that represents the sum of the static pressure and the elevation heads P/rg z, is called the hydraulic grade line. The line that represents the total head of the fluid, P/rg 2/2g z, is called the energy grade
line. The difference between the heights of EGL and HGL is equal to the dynamic head, 2/2g. We note the following about the HGL and EGL:
• For stationary bodies such as reservoirs or lakes, the EGL and HGL
coincide with the free surface of the liquid. The elevation of the free
surface z in such cases represents both the EGL and the HGL since the
velocity is zero and the static pressure (gage) is zero.
533
CHAPTER 12
z
0
12/2g
HGL
EGL
1
Diffuser
22/2g
2
3
Arbitrary reference plane (z = 0)
FIGURE 12–20
The hydraulic grade line (HGL) and the energy grade line (EGL) for free
discharge from a reservoir through a horizontal pipe with a diffuser. At point 0
(at the liquid surface), EGL and HGL are even with the liquid surface since there
is no flow there. HGL decreases rapidly as the liquid accelerates into the pipe;
however EGL decreases very slowly through the well-rounded pipe inlet. EGL
declines continually along the flow direction due to friction and other
irreversible losses in the flow. EGL cannot increase in the flow direction unless
energy is supplied to the fluid. HGL can rise or fall in the flow direction, but can
never exceed EGL. HGL rises in the diffuser section as the velocity decreases,
and the static pressure recovers somewhat; the total pressure does not recover,
however, and EGL decreases through the diffuser. The difference between EGL
and HGL is 21 /2g at point 1, and 22 /2g at point 2. Since 1 2, the difference
between the two grade lines is larger at point 1 than at point 2. The downward
slope of both grade lines is larger for the smaller diameter section of pipe since
the frictional head loss is greater. Finally, HGL decays to the liquid surface at the
outlet since the pressure there is atmospheric. However, EGL is still higher than
HGL by the amount 22 /2g since 3 2 at the outlet.
• The EGL is always a distance 2/2g above the HGL. These two lines
approach each other as the velocity decreases, and they diverge as the
velocity increases. The height of the HGL decreases as the velocity
increases, and vice versa.
(Horizontal)
2/2g
• At a pipe exit, the pressure head is zero (atmospheric pressure) and thus
the HGL coincides with the pipe exit.
• The mechanical energy loss due to frictional effects (conversion to
thermal energy) causes the EGL and HGL to slope downward in the
direction of flow. The slope is a measure of the pipe loss (discussed in
detail in Chap. 14). A component that generates significant frictional
HGL
P
––
ρg
• In an idealized Bernoulli-type flow, EGL is horizontal and its height
remains constant. This would also be the case for HGL when the flow
velocity is constant (Fig. 12–21).
• For open channel flow, the HGL coincides with the free surface of the
liquid, and the EGL is a distance 2/2g above the free surface.
EGL
z
Reference level
0
FIGURE 12–21
In an idealized Bernoulli-type flow,
EGL is horizontal and its height
remains constant. But this is not the
case for HGL when the flow velocity
varies along the flow.
534
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
effects such as a valve causes a sudden drop in both EGL and HGL at
that location.
• A steep jump occurs in EGL and HGL whenever mechanical energy is
added to the fluid (by a pump, for example). Likewise, a steep drop
occurs in EGL and HGL whenever mechanical energy is removed from
the fluid (by a turbine, for example), as shown in Fig. 12–22.
EGL
HGL
Pump
·
Wpump
• The pressure (gage) of a fluid is zero at locations where the HGL
intersects the fluid. The pressure in a flow section that lies above the
HGL is negative, and the pressure in a section that lies below the HGL is
positive (Fig. 12–23). Therefore, an accurate drawing of a piping system
and the HGL can be used to determine the regions where the pressure in
the pipe is negative (below the atmospheric pressure).
Turbine
·
Wturbine
FIGURE 12–22
A steep jump occurs in EGL and HGL
whenever mechanical energy is added
to the fluid by a pump, and a steep
drop occurs whenever mechanical
energy is removed from the fluid
by a turbine.
The last remark enables us to avoid situations in which the pressure drops
below the vapor pressure of the liquid (which causes cavitation, as discussed
in Chap. 10). Proper consideration is necessary in the placement of a liquid
pump to ensure that the suction side pressure does not fall too low, especially at elevated temperatures where vapor pressure is higher than it is at low
temperatures.
12–3
■
APPLICATIONS OF THE
BERNOULLI EQUATION
In Section 12–2, we discussed the fundamental aspects of the Bernoulli equation. In this section, we will demonstrate its use in a wide range of applications through examples.
EXAMPLE 12–3
Spraying Water into the Air
Water is flowing from a hose attached to a water main at 400 kPa gage
(Fig. 12–24). A child places his thumb to cover most of the hose outlet, increasing the pressure upstream of his thumb, causing a thin jet of high-speed
water to emerge. If the hose is held upward, what is the maximum height that
the jet could achieve?
SOLUTION Water from a hose attached to the water main is sprayed into the
air. The maximum height the water jet can rise is to be determined.
Assumptions 1 The flow exiting into the air is steady, incompressible, and irrotational (so that the Bernoulli equation is applicable). 2 The water pressure in
Negative P
P=0
FIGURE 12–23
The pressure (gage) of a fluid
is zero at locations where the HGL
intersects the fluid, and the pressure is
negative (vacuum) in a flow section
that lies above the HGL.
HGL
P=0
Positive P
Positive P
535
CHAPTER 12
2
0
z1
0
P2 22
rg 2g
→
→
P1 21
rg 2g

→
the hose near the outlet is equal to the water main pressure. 3 The surface tension effects are negligible. 4 The friction between the water and air is negligible. 5 The irreversibilities that may occur at the outlet of the hose due to abrupt
expansion are negligible.
Properties We take the density of water to be 1000 kg/m3.
Analysis This problem involves the conversion of flow, kinetic, and potential
energies to each other without involving any pumps, turbines, and wasteful
components with large frictional losses, and thus it is suitable for the use of the
Bernoulli equation. The water height will be maximum under the stated assumptions. The velocity inside the hose is relatively low (1 0) and we take
the hose outlet as the reference level (z1 0). At the top of the water trajectory
2 0, and atmospheric pressure pertains. Then the Bernoulli equation simplifies to
0
z2
→
Water jet
z
1
0
P1 Patm
rg rg z2
Solving for z2 and substituting,
z2 2
P1 Patm P1, gage
1000 N/m2 1 kg m/s
400 kPa
ba
a
rg b
3
2
rg
1N
1 kPa
(1000 kg/m )(9.81 m/s )
Hose
FIGURE 12–24
Schematic for Example 12–3.
40.8 m
Therefore, the water jet can rise as high as 40.8 m into the sky in this case.
Discussion The result obtained by the Bernoulli equation represents the upper
limit and should be interpreted accordingly. It tells us that the water cannot
possibly rise more than 40.8 m, and, in all likelihood, the rise will be much less
than 40.8 m due to irreversible losses that we neglected.
EXAMPLE 12–4
Water Discharge from a Large Tank
A large tank open to the atmosphere is filled with water to a height of 5 m from
the bottom (Fig. 12–25). A tap near the bottom of the tank is now opened, and
water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet.
1
SOLUTION A tap near the bottom of a tank is opened. The exit velocity of water from the tank is to be determined.
Assumptions 1 The flow is incompressible and irrotational (except very close to
the walls). 2 The water drains slowly enough that the flow can be approximated
as steady (actually quasi-steady when the tank begins to drain).
Analysis This problem involves the conversion of flow, kinetic, and potential
energies to each other without involving any pumps, turbines, and wasteful
components with large frictional losses, and thus it is suitable for the use of the
Bernoulli equation. We take point 1 to be at the free surface of water so that
P1 Patm (open to the atmosphere), 1 0 (the tank is large relative to the
outlet), and z1 5 m and z2 0 (we take the reference level at the center of
the outlet). Also, P2 Patm (water discharges into the atmosphere). Then the
Bernoulli equation simplifies to
0
0
P2 22
z1 rg z2
2g
→
→
P1 21
rg 2g
→
z1 22
2g
5m
z
Water
2
0
2
FIGURE 12–25
Schematic for Example 12–4.
536
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Solving for 2 and substituting,
2 22gz1 22(9.81 m/s2)(5 m) 9.9 m/s
The relation 22gz is called the Toricelli equation.
Therefore, the water leaves the tank with an initial velocity of 9.9 m/s. This is
the same velocity that would manifest if a solid were dropped a distance of 5 m
in the absence of air friction drag. (What would the velocity be if the tap were
at the bottom of the tank instead of on the side?)
Discussion If the orifice were sharp-edged instead of rounded, then the flow
would be disturbed, and the velocity would be less than 9.9 m/s, especially near
the edges. Care must be exercised when attempting to apply the Bernoulli equation to situations where abrupt expansions or contractions occur since the friction and flow disturbance in such cases may not be negligible.
EXAMPLE 12–5
z
3
z3
Gasoline
flow intake
tube
2m
Siphoning out Gasoline from a Fuel Tank
During a trip to the beach (Patm 1 atm 101.3 kPa), a car runs out of gasoline, and it becomes necessary to siphon gas out of the car of a good Samaritan (Fig. 12–26). The siphon is a small-diameter hose, and to start the siphon
it is necessary to insert one siphon end in the full gas tank, fill the hose with
gasoline via suction, and then place the other end in a gas can below the level
of the gas tank. The difference in pressure between point 1 (at the free surface
of the gasoline in the tank) and point 2 (at the outlet of the tube) will cause
the liquid to flow from the higher to the lower elevation. Point 2 is located
0.75 m below point 1 in this case, and point 3 is located 2 m above point 1.
The siphon diameter is 4 mm, and frictional losses in the siphon are to be disregarded. Determine (a) the minimum time to withdraw 4 L of gasoline from
the tank to the can and (b) the pressure at point 3. The density of gasoline is
750 kg/m3.
1
2
0
Gas can
FIGURE 12–26
Schematic for Example 12–5.
SOLUTION Gasoline is to be siphoned from a tank. The time it takes to withdraw 4 L of gasoline and the pressure at the highest point in the system are to
be determined.
Assumptions 1 The flow is steady and incompressible. 2 Even though the
Bernoulli equation is not valid through the pipe because of frictional losses, we
employ the Bernoulli equation anyway in order to obtain a best-case estimate.
3 The change in the gasoline surface level inside the tank is negligible compared to elevations z1 and z2 during the siphoning period.
Properties The density of gasoline is given to be 750 kg/m3.
Analysis (a) We take point 1 to be at the free surface of gasoline in the tank so
that P1 Patm (open to the atmosphere), 1 0 (the tank is large relative to
the tube diameter), and z2 0 (point 2 is taken as the reference level). Also,
P2 Patm (gasoline discharges into the atmosphere). Then the Bernoulli equation simplifies to
0
P1 21 →
P2 22
0
z
1
rg 2g
rg 2g z2 →

0.75 m

z1
Gas
tank
→
z1 Solving for 2 and substituting,
2 22gz1 22(9.81 m/s2)(0.75 m) 3.84 m/s
22
2g
537
CHAPTER 12
The cross-sectional area of the tube and the flow rate of gasoline are
A pD 2/4 p(5 103 m)2/4 1.96 105 m2
·
V 2A (3.84 m/s)(1.96 105 m2) 7.53 105 m3/s 0.0753 L/s
Then the time needed to siphon 4 L of gasoline becomes
V
4L
t 53.1 s
V 0.0753 L/s
(b) The pressure at point 3 can be determined by writing the Bernoulli equation
between points 2 and 3. Noting that 2 3 (conservation of mass), z2 0,
and P2 Patm,

→

0 P3 23
P2 22
rg 2g z2 rg 2g z3
→
Patm P3
rg rg z3
Solving for P3 and substituting,
P3 Patm rgz3
1N
1 kPa
101.3 kPa (750 kg/m3)(9.81 m/s2)(2.75 m)a
ba
b
1 kg m/s2 1000 N/m2
81.1 kPa
Discussion The siphoning time is determined by neglecting frictional effects,
and thus this is the minimum time required. In reality, the time will be longer
than 53.1 s because of the friction between the gasoline and the tube surface.
Also, the pressure at point 3 is below the atmospheric pressure. If the elevation
difference between points 1 and 3 is too high, the pressure at point 3 may drop
below the vapor pressure of gasoline at the gasoline temperature, and some
gasoline may evaporate. The vapor then may form a pocket at the top and halt
the flow of gasoline.
EXAMPLE 12–6
Velocity Measurement by a Pitot Tube
A piezometer and a pitot tube are tapped into a horizontal water pipe, as shown
in Fig. 12–27, to measure static and stagnation (static dynamic) pressures.
For the indicated water column heights, determine the velocity at the center of
the pipe.
h3 = 12 cm
SOLUTION The static and stagnation pressures in a horizontal pipe are mea-
h2 = 7 cm
sured. The velocity at the center of the pipe is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Points 1 and 2 are
close enough together that the irreversible energy loss between these two points
is negligible, and thus we can use the Bernoulli equation.
Analysis We take points 1 and 2 along the centerline of the pipe, with point 1
directly under the piezometer and point 2 at the tip of the pitot tube. This is a
steady flow with straight and parallel streamlines, and the gage pressures at
points 1 and 2 can be expressed as
P1 rg(h1 h2)
P2 rg(h1 h2 h3)
Water
1
h1 = 3 cm
2
1
Stagnation
point
FIGURE 12–27
Schematic for Example 12–6.
538
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Noting that point 2 is a stagnation point and thus 2 0 and z1 z2, the
application of the Bernoulli equation between points 1 and 2 gives


0
P1 21
P2 22 →
z2
rg 2g z1 rg 2g
→
21 P2 P1
rg
2g
Substituting the P1 and P2 expressions gives
21 P2 P1 rg(h1 h2 h3) rg(h1 h2)
h3
rg rg
2g
Solving for 1 and substituting,
1 22gh3 22(9.81 m/s2)(0.12 m) 1.53 m/s
Discussion Note that to determine the flow velocity, all we need is to measure
the height of the excess fluid column in the pitot tube.
EXAMPLE 12–7
The Rise of the Ocean Due to Hurricanes
A hurricane is a tropical storm formed over the ocean by low atmospheric
pressures. As a hurricane approaches land, inordinate ocean swells (very high
tides) accompany the hurricane. A Class-5 hurricane features winds in excess
of 155 mph, although the wind velocity at the center “eye” is very low.
Figure 12–28 depicts a hurricane hovering over the ocean swell below. The
atmospheric pressure 200 miles from the eye is 30.0 inHg (at point 1, generally normal for the ocean) and the winds are calm. The hurricane atmospheric
pressure at the eye of the storm is 22.0 inHg. Estimate the ocean swell at
(a) the eye of the hurricane at point 3 and (b) point 2, where the wind velocity
is 155 mph. Take the density of seawater and mercury to be 64 lbm/ft3 and
848 lbm/ft3, respectively, and the density of air at normal sea level temperature
and pressure to be 0.076 lbm/ft3.
SOLUTION A hurricane is moving over the ocean. The amount of ocean swell
at the eye and at active regions of the hurricane are to be determined.
Eye
Hurricane
A
2
Calm
ocean
level
1
3
h2
Ocean
FIGURE 12–28
Schematic for Example 12–7. The
vertical scale is exaggerated.
B
h1
539
CHAPTER 12
Assumptions 1 The airflow within the hurricane is steady, incompressible, and
irrotational (so that the Bernoulli equation is applicable). (This is certainly a
very questionable assumption for a highly turbulent flow, but it is justified in
the solution.) 2 The effect of water drifted into the air is negligible.
Properties The densities of air at normal conditions, seawater, and mercury are
given to be 0.076 lbm/ft3, 64 lbm/ft3, and 848 lbm/ft3, respectively.
Analysis (a) Reduced atmospheric pressure over the water causes the water to
rise. Thus, decreased pressure at point 2 relative to point 1 causes the ocean
water to rise at point 2. The same is true at point 3, where the storm air velocity is negligible. The pressure difference given in terms of the mercury column
height can be expressed in terms of the seawater column height by
rHg
P (rgh)Hg (rgh)sw
→
hsw r hHg
sw
Then the pressure difference between points 1 and 2 in terms of the seawater
column height becomes
rHg
848 lbm/ft3
1 ft
h1 r h Hg a
b[(30 22) inHg]a
b 8.83 ft
sw
12 in
64 lbm/ft3
which is equivalent to the storm surge at the eye of the hurricane since the wind
velocity there is negligible and there are no dynamic effects.
(b) To determine the additional rise of ocean water at point 2 due to the high
winds at that point, we write the Bernoulli equation between points A and B,
which are on top of the points 2 and 3, respectively. Noting that B 0 (the eye
region of the hurricane is relatively calm) and zA zB (both points are on the
same horizontal line), the Bernoulli equation simplifies to


0
PB 2B →
PA 2A
zB
rg 2g zA rg 2g
→
PB PA 2A
rg 2g
Substituting,
PB PA 2A (155 mph)2 1.4667 ft/s 2
rg 2g 2(32.2 ft/s2) a 1 mph b 803 ft
where r is the density of air in the hurricane. Noting that the density of an ideal
gas at constant temperature is proportional to absolute pressure and the density
of air at the normal atmospheric pressure of 14.7 psia 30 inHg is 0.076
lbm/ft3, the density of air in the hurricane is
rair 22 inHg
Pair
r
a
b(0.076 lbm/ft3) 0.056 lbm/ft3
Patm air atm air
30 inHg
Using the relation developed above in part (a), the seawater column height
equivalent to 803 ft of air column height is determined to be
rair
0.056 lbm/ft3
hdynamic r hair a
b(803 ft) 0.70 ft
sw
64 lbm/ft3
Therefore, the pressure at point 2 is 0.70 ft seawater column lower than the
pressure at point 3 due to the high wind velocities, causing the ocean to rise an
additional 0.70 ft. Then the total storm surge at point 2 becomes
h2 h1 hdynamic 8.83 0.70 9.53 ft
540
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Discussion This problem involves highly turbulent flow and the intense breakdown of the streamlines, and thus the applicability of the Bernoulli equation in
part (b) is questionable. The Bernoulli analysis can be thought of as the limiting, ideal case, and shows that the rise of seawater due to high-velocity winds
cannot be more than 0.70 ft.
The wind power of hurricanes is not the only cause of damage to coastal
areas. Ocean flooding and erosion from excessive tides is just as serious, as are
high waves generated by the storm turbulence and energy.
EXAMPLE 12–8
Bernoulli Equation for Compressible Flow
Derive the Bernoulli equation when the compressibility effects are not negligible for an ideal gas undergoing (a) an isothermal process and (b) an isentropic
process.
SOLUTION The Bernoulli equation for compressible flow is to be obtained for
an ideal gas for isothermal and isentropic processes.
Assumptions 1 The flow is steady, and frictional effects are negligible. 2 The
fluid is an ideal gas, so the relation P rRT is applicable. 3 The specific heats
are constant so that P/r k constant during an isentropic process.
Analysis (a) When the compressibility effects are significant and the flow
cannot be assumed to be incompressible, the Bernoulli equation is given by
Eq. 12–17 as
dPr 2 gz constant
2
(1)
The compressibility effects can be properly accounted for by expressing r in
terms of pressure, and then performing the integration dP/r in Eq. (1). But
this requires a relation between P and r for the process. For the isothermal expansion or compression of an ideal gas, the integral in Eq. (1) can be performed
easily by noting that T constant and substituting r P/RT. It gives
dP
RT ln P
dPr P/RT
Substituting into Eq. (1) gives the desired relation,
Isothermal process:
RT ln P 2
gz constant
2
(2)
(b) A more practical case of compressible flow is the isentropic flow of ideal
gases through equipment that involves high-speed fluid flow such as nozzles,
diffusers, and the passages between turbine blades. Isentropic (i.e., reversible
and adiabatic) flow is closely approximated by these devices, and it is characterized by the relation P/r k C constant where k is the specific heat ratio of the gas. Solving for r from P/r k C gives r C 1/k P 1/k. Performing the
integration,
dPr C
1/k
P 1/k dP C 1/k
k
P1/k1
P 1/k P1/k1
P
r
a
b
1/k 1
1/k 1
k1 r
(3)
541
CHAPTER 12
Substituting, the Bernoulli equation for steady, isentropic, compressible flow of
an ideal gas becomes
k
P 2
a
br
gz constant
2
k1
(4a)
P1 21
P2 22
k
k
gz2
br br gz1 a
2
2
k1 1
k1 2
(4b)
Isentropic flow:
or
a
A common practical situation involves the acceleration of a gas from rest (stagnation conditions at state 1) with negligible change in elevation. In that case
we have z1 z2, 1 0. Noting that r P/RT for ideal gases, P/rk constant
for isentropic flow, and the Mach number is defined as Ma /c where
c 2kRT is the local speed of sound for ideal gases, the relation above simplifies to
k /(k1)
P1
k1
c1 a
b Ma 22 d
P2
2
(4c)
where state 1 is the stagnation state and state 2 is any state along the flow.
Discussion It can be shown that the results obtained using the compressible
and incompressible equations deviate no more than 2 percent when the Mach
number is less than 0.3. Therefore, the flow of an ideal gas can be considered
to be incompressible when Ma 0.3. For atmospheric air at normal conditions,
this corresponds to a flow speed of about 100 m/s or 360 km/h, which covers
our range of interest.
12–4
■
ENERGY ANALYSIS OF STEADY-FLOW
SYSTEMS
For steady-flow systems, the time rate of change of the energy content of the
control volume is zero. When the work associated with electric, magnetic, surface tension, and viscous effects are negligible, the steady-flow energy equation can be expressed as (see Chap. 5)
Qnet in Wshaft, net in m ah 2 gzb m ah 2 gzb
2
out
2
(12–26)
2
m⋅ h1 + 1 + gz1
2
In
1
Fixed
control
volume
in
It states that the net rate of energy transfer to a control volume by heat and
work transfers during steady flow is equal to the difference between the rates
of outgoing and incoming energy flows with mass.
Many practical problems involve just one inlet and one outlet (Fig. 12–29).
The mass flow rate for such single-stream systems remains constant, and Eq.
12–26 reduces to
2
2
ah h 2 1 g(z z )b
Qnet in Wshaft, net in m
2
1
2
1
2
(12–27)
Out
2
⋅
⋅
Qnet in + Wshaft, net in
2
m⋅ h2 + 2 + gz2
2
FIGURE 12–29
A control volume with only one inlet
and one outlet and energy interactions.
542
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
where subscripts 1 and 2 stand for inlet and outlet, respectively. The steadyflow energy equation on a unit mass basis is obtained by dividing Eq. 12–27
by the mass flow rate m ,
qnet in wshaft, net in h2 h1 22 21
g(z2 z1)
2
(12–28)
is the net heat transfer to the fluid per unit mass and
where qnet in Qnet in / m
wshaft, net in Wshaft, net in / m is the work input to the fluid per unit mass. Using
the definition of enthalpy h u P/r and rearranging, the steady-flow energy equation can also be expressed as
P1 21
P2
gz1 r gz2 (u2 u1 qnet in)
wshaft, net in r 1
2
2
(12–29)
where u is the internal energy, P/r is the flow energy, 2 /2 is the kinetic energy, and gz is the potential energy of the fluid per unit mass. These relations
are valid for both compressible or incompressible flows.
The left side of Eq. 12–29 represents the mechanical energy input while the
first three terms on the right side represent the mechanical energy output. If
the flow is ideal with no irreversibilities such as friction, the total mechanical
energy must be conserved, and the terms in parentheses (u2 u1 qnet in)
must equal zero. That is,
Ideal flow (no mechanical energy loss)
0.7 kg/s
15.2°C
2 kW
ηpump = 0.70
15.0°C
Water
FIGURE 12–30
The lost mechanical energy in a fluid
flow system results in an increase in
the internal energy of the fluid, and
thus in a rise of fluid temperature.
(12–30)
Any increase in u2 u1 above qnet in is due to the irreversible conversion of
mechanical energy to thermal energy, and thus u2 u1 qnet in represents the
mechanical energy loss (Fig. 12–30). That is,
Mechanical energy loss:
∆u = 0.84 kJ/kg
∆T = 0.2°C
qnet in u2 u1
emech, loss u2 u1 qnet in
(12–31)
For single-phase fluids (a gas or a liquid), we have u2 u1 Cυ(T2 T1)
where Cυ is the constant-volume specific heat. Then the steady-flow energy
equation on a unit mass basis can be written conveniently as a mechanical
energy balance as
emech, in emech, out emech, loss
(12–32)
P1 21
P2 22
wshaft, net in r gz1 r gz2 emech, loss
1
2
2
2
(12–33)
or
Noting that wshaft, net in wshaft, in wshaft, out wpump wturbine, the mechanical
energy balance can be written more explicitly as
P1 21
P2 22
r1 2 gz1 wpump r2 2 gz2 wturbine emech, loss
(12–34)
where wpump is the mechanical work input (due to the presence of a pump, fan,
compressor, etc.) and wturbine is the mechanical work output. When the flow is
incompressible, either absolute or gage pressure can be used for P since Patm/r
would appear on both sides, and would cancel out.
543
CHAPTER 12
gives
Multiplying Eq. 12–34 by the mass flow rate m
2
P1 21
aP2 2 gz b W
gz1b Wpump m
m a r (12–35)
2
turbine Emech, loss
r2
1
2
2
where Wpump is the shaft power input through the pump’s shaft, Wturbine is the
shaft power output through the turbine’s shaft, and E mech, loss is the total mechanical power loss, which consists of pump and turbine losses as well as the
frictional losses in the piping network. That is, E mech, loss E mech, loss, pump E mech, loss, turbine E mech, loss, piping. The energy equation can also be expressed in
terms to heads as
P1 21
P2 22
r1g 2g z1 hpump, u r2g 2g z2 hturbine, e hL
(12–36)
where
wpump, u Wpump, u
is the useful head delivered to the fluid by the pump,
g
mg
wturbine, e Wturbine, e
hturbine, e is the extracted head from the fluid by the turbine,
g
mg
emech, loss, piping Emech, loss, piping
is the irreversible head loss between 1 and 2.
hL g
mg
hpump, u The pump head is zero if the piping system does not involve a pump, a fan, or
a compressor, the turbine head is zero if the system does not involve a turbine,
and both are zero if the flow section analyzed does not involve any mechanical work producing or consuming devices. Also, the head loss is zero in an
idealized situation that does not involve any frictional losses.
Special Case: Incompressible Flow with No
Mechanical Work Devices and Negligible Friction
When the frictional effects are negligible, there is no dissipation of mechanical energy into thermal energy, and thus hL emech, loss, piping/g 0, as shown
in Example 12–9. Also, hpump, u hturbine, e 0 when there are no mechanical
work devices such as fans, pumps, or turbines, and r1 r2 r when the fluid
is incompressible. Then equation 12–36 reduces to (Fig. 12–31)
P1 21
P2 22
z
1
rg 2g
rg 2g z2
or
P 2
rg 2g z constant
(12–37)
which is the Bernoulli equation derived earlier using Newton’s second law of
motion.
Energy equation:
0
0
0
P1 12
P2 22
––
+
+
z
––
+
h
=
––
+
––
+
z
+
h
+
h
1
pump, u
2
turbine, e
L
ρg 2g
ρg 2g
Bernoulli equation:
P2 22
P1 12
–– + z2
–– + z1 = ρ––
ρ––
g + 2g
g + 2g
FIGURE 12–31
When the frictional losses are
negligible and there are no work
devices involved, the energy equation
for steady incompressible flow
reduces to the Bernoulli equation.
544
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Kinetic Energy Correction Factor, a
(r)
A
m· = ρ m A, ρ = constant
∫
·
KEact = keδm· =
∫ ––21 2 (r)[ ρ(r)
A
dA]
∫
1 ρ 3(r) dA
= ––
A
2
·
1 m· 2 = ––
1 ρ A3
KEm = ––
m
m
2
2
·
KEact
(r)
1
–––
α = ––––––
= ––
·
A A m
KEm
∫
3
( )
dA
FIGURE 12–32
The determination of the kinetic
energy correction factor using actual
velocity distribution (r) and the
mean velocity m at a cross section.
The mean flow velocity m was defined such that the relation rmA gives the
actual mass flow rate. Therefore, there is no such thing as a correction factor
for mass flow rate. However, the kinetic energy of a fluid stream obtained
from 2/2 is not the same as the actual kinetic energy of the fluid stream since
the square of a sum is not equal to the sum of the squares of its components
(Fig. 12–32). This error can be corrected by replacing the kinetic energy terms
2/2 in the energy equation by a2m /2 where a is the kinetic energy correction factor. By using equations for the variation of velocity with the radial
distance, it can be shown that the correction factor is 2.0 for fully developed
laminar pipe flow, and it ranges between 1.04 and 1.11 for turbulent flow in a
circular pipe.
The kinetic energy correction factors are usually disregarded in an elementary analysis since (1) most flows encountered in practice are turbulent, for
which the correction factor is near unity, and (2) the kinetic energy terms are
usually small relative to the other terms in the energy equation, and multiplying them by a factor less than 2.0 does not make much difference. Besides,
when the velocity and thus the kinetic energy are high, the flow turns turbulent. Therefore, we will not consider the kinetic energy correction factor in the
analysis. However, the reader should keep in mind that he or she may encounter situations for which these factors are significant, especially when the
flow is laminar.
EXAMPLE 12–9
Effect of Friction on Fluid Temperature and
Head Loss
Show that during steady and incompressible flow of a fluid in an adiabatic flow
section (a) the temperature remains constant and there is no head loss when
the flow is frictionless and (b) the temperature increases and some head loss
occurs when there are frictional effects. Discuss if it is possible for the fluid
temperature to decrease during such flow (Fig. 12–33).
1
T1
u1
ρ = constant
(adiabatic)
FIGURE 12–33
Schematic for Example 12–9.
2
T2
u2
SOLUTION Steady and incompressible flow through an adiabatic section is
considered. The effects of friction on the temperature and the heat loss are to
be determined.
Assumptions 1 The flow is steady and incompressible. 2 The flow section is
adiabatic and thus there is no heat transfer.
Analysis The density of a fluid remains constant during incompressible flow,
and the entropy change of an incompressible system is
s Cυ ln
T2
T1
This relation represents the entropy change of the fluid per unit mass as it flows
through the flow section from state 1 at the inlet to state 2 at the exit. Entropy
change is caused by two effects: (1) heat transfer and (2) irreversibilities.
Therefore, in the absence of heat transfer, entropy change is due to irreversibilities only whose effect is always to increase entropy.
(a) The entropy change of the fluid is zero when the process does not involve
any irreversibilities such as friction and swirling, and thus for reversible flow
we have
545
CHAPTER 12
T2
0
→
T2 T1
T1
emech, loss u2 u1 Cυ(T2 T1) 0
s Cυ ln
Temperature change:
Mechanical energy loss:
hL emech, loss /g 0
Head loss:
Thus we conclude that when heat transfer and frictional effects are negligible,
(1) the temperature of the fluid remains constant, (2) no mechanical energy is
converted to thermal energy, and (3) there is no head loss.
(b) When there are irreversibilities such as friction, the entropy change is positive and thus we have:
T2
0
→
T2 T1
T1
emech, loss u2 u1 Cυ (T2 T1) 0
hL emech, loss /g 0
s Cυ ln
Temperature change:
Mechanical energy loss:
Head loss:
Thus we conclude that when the flow is adiabatic and irreversible, (1) the temperature of the fluid increases, (2) some mechanical energy is converted to
thermal energy, and (3) some head loss occurs.
Discussion It is impossible for the fluid temperature to decrease during
steady, incompressible, adiabatic flow since this would require the entropy of an
adiabatic system to decrease, which would be a violation of the second law of
thermodynamics.
EXAMPLE 12–10
Pumping Power and Frictional Heating in a
Pump
The pump of a water distribution system is powered by a 15-kW electric motor
whose efficiency is 90 percent (Fig. 12–34). The water flow rate through the
pump is 50 L/s. The diameters of the inlet and exit pipes are the same, and the
elevation difference across the pump is negligible. If the pressures at the inlet
and outlet of the pump are measured to be 100 kPa and 300 kPa (absolute),
respectively, determine (a) the mechanical efficiency of the pump and (b) the
temperature rise of water as it flows through the pump due to the mechanical
inefficiency.
Water
50 L/s
300 kPa
η motor = 90%
·
Wpump
SOLUTION The pressures across a pump are measured. The mechanical efficiency of the pump and the temperature rise of water are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The pump is driven by
an external motor so that the heat generated by the motor is dissipated to the
atmosphere. 3 The elevation difference between the inlet and outlet of the
pump is negligible, z1 z2. 4 The inlet and outlet diameters are the same and
thus the inlet and exit velocities are equal, 1 2.
Properties We take the density of water to be 1 kg/L 1000 kg/m3 and its
specific heat to be 4.18 kJ/kg · C.
Analysis (a) The mass flow rate of water through the pump is
m rV (1 kg/L)(50 L/s) 50 kg/s
The motor draws 15 kW of power and is 90 percent efficient. Thus the mechanical (shaft) power it delivers to the pump is
Wpump, shaft hmotor Welectric (0.90)(15 kW) 13.5kW
2
Motor
15 kW
100 kPa
1
FIGURE 12–34
Schematic for Example 12–10.
546
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
To determine the mechanical efficiency of the pump, we need to know the
increase in the mechanical energy of the fluid as it flows through the pump,
which is
P1 21
P2 22
Emech, fluid Emech, out Emech, in m a r gz2b m a r gz1b
2
2
Simplifying it for this case and substituting the given values,
aP2 P1b (50 kg/s)a(300 100) kPaba 1 kJ b 10 kW
Emech, fluid m
r
1000 kg/m3
1 kPa m3
Then the mechanical efficiency of the pump becomes
Emech, fluid
10 kW
0.741
hpump 13.5kW
Wpump, shaft
or 74.1%
(b) Of the 13.5-kW mechanical power supplied by the pump, only 10 kW is imparted to the fluid as mechanical energy. The remaining 3.5 kW is converted to
thermal energy due to frictional effects, and this “lost” mechanical energy manifests itself as a heating effect in the fluid,
Emech, loss W pump, shaft Emech, fluid 13.5 10 3.5 kW
The temperature rise of water due to this mechanical inefficiency is determined
(u u ) m
CT. Solving for T,
from the thermal energy balance, E mech, loss m
2
1
T Emech, loss
3.5 kW
0.017C
(50 kg/s)(4.18 kJ/kg C)
mC
Therefore, the water will experience a temperature rise of 0.017C, which is
very small, as it flows through the pump.
Discussion In an actual application, the temperature rise of water will probably be less since part of the heat generated will be transferred to the casing of
the pump, and from the casing to the surrounding air. If the entire pump motor
were submerged in water, then the 1.5 kW dissipated to the air due to motor inefficiency would also be transferred to the surrounding water as heat. This
would cause the water temperature to rise more.
1
EXAMPLE 12–11
100 m3/s
120 m
hL = 35m
Turbine
2
Generator
η turbine-gen = 80%
FIGURE 12–35
Schematic for Example 12–11.
Hydroelectric Power Generation from a Dam
In a hydroelectric power plant, 100 m3/s of water flows from an elevation of
120 m to a turbine, where electric power is generated (Fig. 12–35). The total
head loss in the system from point 1 to point 2 (excluding the turbine unit) is
determined to be 35 m. If the overall efficiency of the turbine-generator is 80
percent, estimate the electrical power output.
SOLUTION The available head, flow rate, head loss, and efficiency of a hydroelectric turbine are given. The electrical power output is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The water levels at the
reservoir and the discharge site remain constant.
547
CHAPTER 12
Properties We take the density of water to be 1000 kg/m3.
Analysis The mass flow rate of water through the turbine is
m r V (1000 kg/m3)(100 m3/s) 105 kg/s
We take point 2 as the reference level, and thus z2 0. Also, both points 1 and
2 are open to the atmosphere (P1 P2 Patm) and the flow velocities are negligible at both points (1 2 0). Then the energy equation for steady incompressible flow reduces to




P1 21
P
2
→u0 rg1 2 z2 →0 hturbine, e hL → hturbine, e z1 hL
rg 2g z1 hpump,
2g
Substituting, the extracted turbine head and the corresponding turbine power
are determined to be
hturbine, e z1 hL 120 35 85 m
1 kJ/kg
gh
5
2
W turbine, e m
b 83,400 kW
turbine, e (10 kg/s)(9.81 m/s )(85 m)a
1000 m2/s2
Therefore, a perfect turbine-generator would generate 83,400 kW of electricity
from this resource. The electric power generated by the actual unit is
W electric hturbine-gen W turbine, e (0.80)(83.4 MW) 66.7 MW
Discussion Note that the power generation will increase by almost 1 MW for
each percentage point improvement in the efficiency of the turbine-generator
unit.
EXAMPLE 12–12
Fan Selection for Air Cooling of a Computer
A fan is to be selected to cool a computer case whose dimensions are 12 cm 40 cm 40 cm (Fig. 12–36). Half of the volume in the case is expected to be
filled with components and the other half to be air space. A 6-cm-diameter hole
is available at the front of the case for the installation of the fan that is to replace the air in the void spaces of the case once every second. Small low-power
fan-motor-combined units are available in the market and their efficiency is estimated to be 30 percent. Determine (a) the wattage of the fan-motor unit to be
purchased and (b) the pressure difference across the fan. Take the air density
to be 1.20 kg/m3.
SOLUTION A fan is to cool a computer case by completely replacing the air inside once every second. The power of the fan and the pressure difference across
it are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 Losses other than
those due to the inefficiency of the fan-motor unit are negligible (hL 0).
Properties The density of air is given to be 1.20 kg/m3.
Analysis (a) Noting that half of the volume of the case is occupied by the
components, the air volume in the computer case is
V (Void fraction)(Total case volume)
0.5(12 cm 40 cm 40 cm) 9600 cm3
Streamline
3
4
1
·
Welect
2
2
Fan
FIGURE 12–36
Schematic for Example 12–12.
548
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Therefore, the volume and mass flow rates of air through the case are
9600 cm3
V
9600 cm3/s 9.6 103 m3/s
V 1s
t
r V (1.20 kg/m3)(9.6 103 m3/s) 0.0115 kg/s
m
The cross-sectional area of the opening in the case and the average air velocity are
pD 2 p(0.06 m)2
2.83 103 m2
4
14
V 9.6 103 m3/s
3.39 m/s
A 2.83 103 m2
A
We draw the control volume around the fan such that both the inlet and the exit
are at the atmospheric pressure (P1 P2 Patm), as shown in the figure, and
the inlet section 1 is large and far from the fan so that the flow velocity at the
inlet section is negligible (1 0). Noting that z1 z2 and the mechanical
losses due to fan inefficiency and the frictional effects in flow are disregarded,
the energy equation simplifies to


0
2
→
P1 21 
→0 
0
→0


aP2 2 gz →0 b W

mar gz1→ b W fan m
turbine Emech, loss
2
r
2
2
Solving for W fan and substituting,
2
2
2 (0.0115 kg/s) (3.39 m/s) a 1 N b 0.066 W
W fan m
2
2
1 kg m/s2
Then the required electrical power input to the fan is determined to be
Wfan
W elect h
fan-motor
0.066 W
0.22 W
0.3
Therefore, a fan-motor rated at 0.22 W is adequate for this job.
(b) To determine the pressure difference across the fan unit, we take points
3 and 4 to be on the two sides of the fan on a horizontal line. This time again
z3 z4 and 3 4 since the fan is a narrow cross section. Disregarding mechanical losses, the energy equation reduces to
P3
→0
P4 E 




m r W fan m
mech, loss
r
→
P4 P3
W fan m
r
Solving for P4 P3 and substituting,
P4 P3 r Wfan (1.2 kg/m3)(0.066 W) 1 Pa m3
b 6.9 Pa
a
1 Ws
0.0115 kg/s
m
Therefore, the pressure rise across the fan is 6.9 Pa.
Discussion The efficiency of the fan-motor unit is given to be 30 percent,
·
which means 30 percent of the electric power W electric consumed by the unit will
be converted to useful mechanical energy while the rest (70 percent) will be
“lost” and converted to thermal energy. Also, a much larger fan is required in an
actual system to overcome frictional losses inside the computer case.
549
CHAPTER 12
EXAMPLE 12–13
Head and Power Loss during Water Pumping
Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20 kW of useful mechanical power to water (Fig. 12–37). The free surface
of the upper reservoir is 45 m higher than the surface of the lower reservoir. If
the flow rate of water is measured to be 0.03 m3/s, determine the head loss of
the system and the lost mechanical power during this process.
SOLUTION Water is pumped from a lower reservoir to a higher one. The head
and power loss associated with this process are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the reservoirs is constant.
Properties We take the density of water to be 1000 kg/m3.
Analysis The mass flow rate of water through the system is
m r V (1000 kg/m3)(0.03 m3/s) 30 kg/s
We choose points 1 and 2 at the free surfaces of the lower and upper reservoirs,
respectively, and take the surface of the lower reservoir as the reference level
(z1 0). Both points are open to the atmosphere (P1 P2 Patm) and the velocities at both locations are negligible (1 2 0). Then the energy equation for steady incompressible flow for a control volume between 1 and 2
reduces to
2
0.03 m3/s
45 m
1 z1 = 0
20 kW
Pump
System
boundary
FIGURE 12–37
Schematic for Example 12–13.


0
0
P1 21 
P2 22 →
→
→0 0
m a r gz1→ b W pump m a r  gz2 b W 
turbine Emech, loss
2
2
W pump m gz2 Emech, loss
Emech, loss W pump m gz2
→
Substituting, the lost mechanical power and head loss are determined to be
1 kW
1N
ba
Emech, loss 20 kW (30 kg/s)(9.81 m/s2)(45 m)a
b
1 kg m/s2 1000 N m/s
6.76 kW
1 kg m/s2 1000 N m/s
Emech, loss
6.76 kW
b 23.0 m
a
ba
hL 1N
1 kW
(30 kg/s)(9.81 m/s2)
mg
since the entire mechanical losses are due to frictional losses in the pipes.
Discussion The 6.76 kW of power is used to overcome the friction in the piping system. Note that the pump could raise the water an additional 23 m if
there were no losses in the system. In this ideal case, the pump would function
as a turbine when the water is allowed to flow from the upper reservoir to the
lower reservoir, and generate 20 kW of power.
SUMMARY
This chapter deals with the Bernoulli and the energy equations,
and their applications. The mechanical energy is the form of
energy associated with the velocity, elevation, and pressure of
the fluid, and it can be converted to mechanical work completely and directly by an ideal mechanical device. The efficiencies of various devices are defined as
550
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Mechanical energy increase of the fluid
Mechanical energy input
Emech, fluid Wpump, u
Wpump
Wshaft, in
hpump Mechanical energy output
Mechanical energy decrease of the fluid
Wshaft, out
Wturbine
Emech, fluid Wturbine, e
Mechanical power output Wshaft, out
hmotor Electrical power input
Welect, in
Welect, out
Electrical power output
hgenerator Mechanical power input W shaft, in
Emech, out Emech, in
hpump-motor hpump hmotor Welect, in
Emech, fluid
Welect, in
hturbine hturbine-gen hturbine hgenerator
Welect, out
Welect, out
Emech, in Emech, out Emech, fluid
The Bernoulli equation is a relation between pressure, velocity,
and elevation in steady incompressible flow, and is expressed
along a streamline and in regions where net viscous forces are
negligible as
P 2
r 2 gz constant
sents the pressure rise when the fluid in motion is brought to a
stop; and rgz is the hydrostatic pressure, which accounts for
the effects of fluid weight on pressure. The sum of the static,
dynamic, and hydrostatic pressures is called the total pressure.
The Bernoulli equation states that the total pressure along a
streamline is constant. The sum of the static and dynamic pressures is called the stagnation pressure, which represents the
pressure at a point where the fluid is brought to a complete stop
in a frictionless manner. The Bernoulli equation can also be
represented in terms of “heads” by dividing each term by g,
P 2
rg 2g z H constant
where P/rg is the pressure head, which represents the height of
a fluid column that produces the static pressure P; 2/2g is the
velocity head, which represents the elevation needed for a fluid
to reach the velocity during frictionless free fall; and z is the
elevation head, which represents the potential energy of the
fluid. Also, H is the total head for the flow. The line that represents the sum of the static pressure and the elevation heads,
P/rg z, is called the hydraulic grade line (HGL), and the line
that represents the total head of the fluid, P/rg 2/2g z, is
called the energy grade line (EGL).
The energy equation for steady incompressible flow can be
expressed as
P1 21
P2 22
r 2 gz1 wpump r 2 gz2
wturbine emech, loss
P2 22
P1 21
m a r gz1b W pump m a r gz2b
2
2
W turbine Emech, loss
P2 22
P1 21
rg 2g z1 hpump, u rg 2g z2 hturbine, e hL
It can also be expressed between any two points on the streamline as
P2 22
P1 21
r 2 gz1 r 2 gz2
The Bernoulli equation is an expression of mechanical energy
balance and can be stated as the sum of the kinetic, potential,
and flow energies of a fluid particle is constant along a streamline during steady flow when the compressibility and frictional
effects are negligible. Multiplying the Bernoulli equation by
density gives
Pr
2
rgz constant
2
where P is the static pressure, which represents the actual pressure of the fluid; r2/2 is the dynamic pressure, which repre-
where
wpump, u Wpump, u
hpump, u g mg
hturbine, e hL wturbine, e Wturbine, e
g
mg
emech loss, piping Emech loss, piping
g
mg
emech, loss u2 u1 qnet in
551
CHAPTER 12
REFERENCES AND SUGGESTED READING
1. C. T. Crowe, J. A. Roberson, and D. F. Elger. Engineering
Fluid Mechanics. 7th ed. New York: Wiley, 2001.
4. Panton, R. L. Incompressible Flow. 2nd ed. New York:
Wiley, 1996.
2. R. C. Dorf, ed. in chief. The Engineering Handbook. Boca
Raton, FL: CRC Press, 1995.
5. M. C. Potter and D. C. Wiggert. Mechanics of Fluids.
2nd ed. Upper Saddle River, NJ: Prentice Hall, 1997.
3. B. R. Munson, D. F. Young, and T. Okiishi. Fundamentals
of Fluid Mechanics. 4th ed. New York: Wiley, 2002
6. M. Van Dyke. An Album of Fluid Motion. Stanford, CA:
The Parabolic Press, 1982.
PROBLEMS*
Mechanical Energy and Efficiency
12–1C What is mechanical energy? How does it differ from
thermal energy? What are the forms of mechanical energy of a
fluid stream?
12–2C What is mechanical efficiency? What does a mechanical efficiency of 100 percent mean for a hydraulic turbine?
12–3C How is the combined pump-motor efficiency of a
pump and motor system defined? Can the combined pumpmotor efficiency be greater than either of the pump or the
motor efficiency?
12–4C Define turbine efficiency, generator efficiency, and
combined turbine-generator efficiency.
12–5 Consider a river flowing toward a lake at an average
velocity of 3 m/s at a rate of 500 m3/s at a location 90 m above
the lake surface. Determine the total mechanical energy of the
river water per unit mass and the power generation potential of
Answer: 444 MW
the entire river at that location.
River
3 m/s
90 m
FIGURE P12–5
*Problems designated by a “C” are concept questions, and
students are encouraged to answer them all. Problems designated
by an “E” are in English units, and the SI users can ignore them.
Problems with a CD-EES icon
are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computer-EES icon
are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text.
12–6 Electrical power is to be generated by installing a hydraulic turbine-generator at a site 70 m below the free surface
of a large water reservoir that can supply water at a rate of
1500 kg/s steadily. If the mechanical power output of the turbine is 800 kW and the electrical power generation is 750 kW,
determine the turbine efficiency and the combined turbinegenerator efficiency of this plant. Neglect losses in the pipes.
12–7 At a certain location, wind is blowing steadily at 12 m/s.
Determine the mechanical energy of air per unit mass and the
power generation potential of a wind turbine with 50-mdiameter blades at that location. Also determine the actual electric power generation assuming an overall efficiency of 30
percent. Take the air density to be 1.25 kg/m3.
12–8
Reconsider Prob. 12–7. Using EES (or other)
software, investigate the effect of wind velocity
and the blade span diameter on wind power generation. Let the
velocity vary from 5 m/s to 20 m/s in increments of 5 m/s, and
the diameter to vary from 20 m to 80 m in increments of 20 m.
Tabulate the results, and discuss their significance.
12–9E A differential thermocouple with sensors at the inlet
and exit of a pump indicates that the temperature of water rises
0.072°F as it flows through the pump at a rate of 1.5 ft3/s. If the
shaft power input to the pump is 27 hp, determine the mechanical efficiency of the pump. Answer: 64.7%
∆T = 0.072° F
Pump
FIGURE P12–9E
12–10 Water is pumped from a lake to a storage tank 20 m
above at a rate of 70 L/s while consuming 20.4 kW of electric
power. Disregarding any frictional losses in the pipes and any
changes in kinetic energy, determine (a) the overall efficiency
of the pump-motor unit and (b) the pressure difference between
the inlet and the exit of the pump.
552
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
Storage tank
12–23C A glass manometer with oil as the working fluid is
connected to an air duct as shown in the figure. Will the oil in
the manometer move as in figure (a) or in figure (b)? Explain.
What would your response be if the flow direction is reversed?
20 m
Pump
Flow
Flow
FIGURE P12–10
12–11C What is streamwise acceleration? How does it differ
from normal acceleration? Can a fluid particle accelerate in
steady flow?
12–12C Express the Bernoulli equation in three different
ways using (a) energies, (b) pressures, and (c) heads.
12–13C What are the three major assumptions used in the
derivation of the Bernoulli equation?
FIGURE P12–23C
12–24C The velocity of a fluid flowing in a pipe is to be
measured by two different pitot-type mercury manometers
shown in the figure. Would you expect both manometers to
predict the same velocity for flowing water? If not, which
would be more accurate? Explain. What would your response
be if air were flowing in the pipe instead of water?
Flow
12–14C Define static, dynamic, and hydrostatic pressure.
Under what conditions is their sum constant for a flow stream?
12–15C What is stagnation pressure? Explain how it can be
measured.
1
12–16C Define pressure head, velocity head, and elevation
head for a fluid stream and express them for a fluid stream
whose pressure is P, velocity is , and elevation is z.
12–17C What is the hydraulic grade line? How does it differ
from the energy grade line? Under what conditions do both
lines coincide with the free surface of a liquid?
12–18C How is the location of the hydraulic grade line determined for open-channel flow? How is it determined at the
exit of a pipe discharging to the atmosphere?
12–19C The water level of a tank on a building roof is 20 m
above the ground. A hose leads from the tank bottom to the
ground. The end of the hose has a nozzle, which is pointed
straight up. What is the maximum height to which the water
could rise? What factors would reduce this height?
12–20C In a certain application, a siphon must go over a
high wall. Can water or oil with a specific gravity of 0.8 go
over a higher wall? Why?
12–21C Explain how and why a siphon works. Someone
proposes siphoning cold water over a 7-m-high wall. Is this
feasible? Explain.
12–22C A student siphons water over a 8.5-m wall at sea
level. He then climbs to the summit of Mount Shasta (elevation
4390 m, Patm 58.5 kPa) and attempts the same experiment.
Comment on his prospects for success.
(b)
(a)
Bernoulli Equation
2
FIGURE P12–24
12–25 In cold climates, the water pipes may freeze and burst
if proper precautions are not taken. In such an occurrence, the
exposed part of a pipe on the ground ruptures, and water shoots
up to 34 m. Estimate the gage pressure of water in the pipe.
State your assumptions and discuss if the actual pressure is
more or less than the value you predicted.
12–26 A pitot tube is used to measure the velocity of an aircraft flying at 3000 m. If the differential pressure reading is
3 kPa, determine the velocity of the aircraft.
12–27 While traveling on a dirt road, the bottom of a car hits
a sharp rock and a small hole develops at the bottom of its gas
tank. If the height of the gasoline in the tank is 30 cm, determine the initial velocity of the gasoline at the hole. Discuss
how the velocity will change with time and how the flow will
be affected if the lid of the tank is closed tightly.
Answer: 2.43 m/s
12–28E
The drinking water needs of an office are met
by large water bottles. One end of a 0.25-indiameter plastic hose is inserted into the bottle placed on a high
stand, while the other end with an on/off valve is maintained
2 ft below the bottom of the bottle. If the water level in the bottle is 1.5 ft when it is full, determine how long it will take at the
553
CHAPTER 12
minimum to fill an 8-oz glass ( 0.00835 ft3) (a) when the bottle is first opened and (b) when the bottle is almost empty.
1.5 ft
Water
2 ft
12–32
Reconsider Prob. 12–31. Using EES (or other)
software, investigate the effect of water height in
the tank on the discharge velocity. Let the water height vary
from 0 to 5 m in increments of 0.5 m. Tabulate and plot the
results.
12–33E A siphon pumps water from a large reservoir to a
lower tank that is initially empty. The tank also has a rounded
orifice 15 ft below the reservoir surface where the water leaves
the tank. Both the siphon and the orifice diameters are 2 in. Ignoring frictional losses, determine to what height the water will
rise in the tank at equilibrium.
12–34 Water enters a tank of diameter DT steadily at a mass
. An orifice at the bottom with diameter D alflow rate of m
in
o
lows water to escape. The orifice has a rounded entrance, so
the frictional losses are negligible. If the tank is initially empty,
(a) determine the maximum height that the water will reach in
the tank and (b) obtain a relation for water height z as a function of time.
FIGURE P12–28E
m⋅
12–29 A piezometer and a pitot tube are tapped into a 3-cmdiameter horizontal water pipe, and the height of the water
columns are measured to be 20 cm in the piezometer and
35 cm in the pitot tube (both measured from the top surface of
the pipe). Determine the velocity at the center of the pipe.
DT
12–30 The diameter of a cylindrical water tank is Do and its
height is H. The tank is filled with water, which is open to the
atmosphere. An orifice of diameter Do with a smooth entrance
(i.e., no losses) is open at the bottom. Develop a relation for the
time required for the tank (a) to empty halfway and (b) to
empty completely.
12–31 A pressurized tank of water has a 10-cm-diameter orifice at the bottom, where water discharges to the atmosphere.
The water level is 3 m above the outlet. The tank air pressure
above the water level is 300 kPa (absolute) while the atmospheric pressure is 100 kPa. Neglecting frictional effects, determine the initial discharge rate of water from the tank.
Answer: 0.168 m3/s
Air
300 kPa
z
Do
FIGURE P12–34
12–35E Water flows through a horizontal pipe at a rate of
1 gal/s. The pipe consists of two sections of diameters 4 in and
2 in with a smooth reducing section. The pressure difference
between the two pipe sections is measured by a mercury
manometer. Neglecting frictional effects, determine the differential height of mercury between the two pipe sections.
Answer: 0.52 in
4 in
2 in
3m
10 cm
FIGURE P12–31
h
FIGURE P12–35E
554
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
12–36 An airplane is flying at an altitude of 12,000 m. Determine the gage pressure at the stagnation point on the nose of
the plane if the speed of the plane is 200 km/h. How would you
solve this problem if the speed were 1050 km/h? Explain.
12–37 The air velocity in the duct of a heating system is to be
measured by a pitot tube inserted into the duct parallel to flow.
If the differential height between the water columns connected
to the two outlets of the pitot tube is 2.4 cm, determine (a) the
flow velocity and (b) the pressure rise at the tip of the pitot
tube. The air temperature and pressure in the duct are 45C and
98 kPa.
12–38 The water in a 10-m-diameter, 2-m-high above-theground swimming pool is to be emptied by unplugging a 3-cmdiameter, 25-m-long horizontal pipe attached to the bottom of
the pool. Determine the maximum discharge rate of water
through the pipe. Also, explain why the actual flow rate will
be less.
12–39 Reconsider Prob. 12–38. Determine how long it will
take to empty the swimming pool completely.
Answer: 19.7 h
12–40
Reconsider Prob. 12–39. Using EES (or other)
software, investigate the effect of the discharge
pipe diameter on the time required to empty the pool completely. Let the diameter vary from 1 to 10 cm in increments of
1 cm. Tabulate and plot the results.
12–41 Air at 110 kPa and 50C flows upward through a
6-cm-diameter inclined duct at a rate of 45 L/s. The duct diameter is then reduced to 4 cm through a reducer. The pressure
change across the reducer is measured by a water manometer.
The elevation difference between the two points on the pipe
where the two arms of the manometer are attached is 0.20 m.
Determine the differential height between fluid levels of the
two arms of the manometer.
and determine the flow rate of air. Take the air density to be
0.075 lbm/ft3.
12.2 psia
Air
11.8 psia
2.6 in
1.8 in
FIGURE P12–42E
12–43 The water pressure in the mains of a city at a particular location is 400 kPa gage. Determine if this main can serve
water to neighborhoods that are 50 m above this location.
12–44 A handheld bicycle pump can be used as an atomizer
to generate a fine mist of paint or pesticide by forcing air at a
high velocity through a small hole and placing a short tube between the liquid reservoir and the high-speed air jet whose low
pressure drives the liquid up through the tube. In such an
atomizer, the hole diameter is 0.3 cm, the vertical distance between the liquid level in the tube and the hole is 10 cm, and the
bore (diameter) and the stroke of the air pump are 5 cm and
20 cm, respectively. If the atmospheric conditions are 20C and
95 kPa, determine the minimum speed that the piston must be
moved in the cylinder during pumping to initiate the atomizing
effect. The liquid reservoir is open to the atmosphere.
20 cm
5 cm
0.3 cm
Air
Liquid
rising
10 cm
FIGURE P12–44
12–45 The water level in a tank is 20 m above the ground. A
hose is connected to the bottom of the tank, and the nozzle at
the end of the hose is pointed straight up. The tank cover is airtight, and the air pressure above the water surface is 2 atm
Air
h
FIGURE P12–41
12–42E Air is flowing through a venturi meter whose diameter is 2.6 in at the entrance part (location 1) and 1.8 in at the
throat (location 2). The gage pressure is measured to be 12.2
psia at the entrance and 11.8 psia at the throat. Neglecting frictional effects, show that the volume flow rate can be expressed as
2(P1 P2)
V A2
B r(1 A22 /A21)
2 atm
h
20 m
FIGURE P12–45
555
CHAPTER 12
gage. The system is at sea level. Determine the maximum
height to which the water stream could rise. Answer: 40.7 m
12–46 A pitot tube connected to a water manometer is used
to measure the velocity of air. If the deflection (the vertical
distance between the fluid levels in the two arms) is 7.3 cm,
determine the air velocity. Take the density of air to be
1.25 kg/m3.
the pipe is 5 cm on the intake side and 7 cm on the discharge
side. Determine (a) the maximum flow rate of water and
(b) the pressure difference across the pump. Assume the
elevation difference between the pump inlet and the outlet to
be negligible.
Pool
30 m
Air
Pitot
tube
7.3 cm
Manometer
FIGURE P12–46
12–47E The air velocity in a duct is measured by a pitot tube
connected to a differential pressure gage. If the air is at 13.4
psia absolute and 70F and the reading of the differential pressure gage is 0.15 psi, determine the air velocity.
FIGURE P12–55
Answer: 142.7 ft/s
12–56 Reconsider Prob. 12–55. Determine the flow rate of
water and the pressure difference across the pump if the head
loss of the piping system is 5 m.
12–48 In a hydroelectric power plant, water enters the turbine nozzles at 700 kPa absolute with a low velocity. If the exit
pressure is the atmospheric pressure of 100 kPa, determine the
maximum velocity to which water can be accelerated by the
nozzles before striking the turbine blades.
12–57E In a hydroelectric power plant, water flows from an
elevation of 240 ft to a turbine, where electric power is generated. For an overall turbine-generator efficiency of 83 percent,
determine the minimum flow rate required to generate 100 kW
of electricity. Answer: 370 lbm/s
Energy Equation
12–58E Reconsider Prob. 12–57E. Determine the flow rate
of water if the head loss of the piping system between the free
surfaces of the source and the sink is 36 ft.
12–49C Consider the steady adiabatic flow of an incompressible fluid. Can the temperature of the fluid decrease during flow? Explain.
12–50C Consider the steady adiabatic flow of an incompressible fluid. If the temperature of the fluid remains constant
during flow, is it accurate to say that the frictional effects are
negligible?
12–51C What is head loss? How is it related to the mechanical energy loss?
12–52C What is pump head? How is it related to the power
input to the pump?
12–59
A fan is to be selected to ventilate a bathroom
whose dimensions are 2 m 3 m 3 m. The
air velocity is not to exceed 8 m/s to minimize vibration and
noise. The combined efficiency of the fan-motor unit to be
used can be taken to be 50 percent. If the fan is to replace the
entire volume of air in 10 min, determine (a) the wattage of the
fan-motor unit to be purchased, (b) the diameter of the fan
casing, and (c) the pressure difference across the fan. Take the
air density to be 1.25 kg/m3.
8 m/s
12–53C What is the kinetic energy correction factor? Is it
significant?
12–54C The water level in a tank is 20 m above the ground.
A hose is connected to the bottom of the tank, and the nozzle at
the end of the hose is pointed straight up. The water stream
from the nozzle is observed to rise 25 m above the ground. Explain what may cause the water from the hose to rise above the
tank level.
12–55 Underground water is to be pumped by a 70 percent
efficient 3-kW submerged pump to a pool whose free surface
is 30 m above the underground water level. The diameter of
Exhaust
fan
Air
FIGURE P12–59
12–60 Water is being pumped from a large lake to a reservoir
25 m above at a rate of 25 L/s by a 10-kW (shaft) pump. If the
head loss of the piping system is 7 m, determine the mechanical efficiency of the pump. Answer: 78.5%
556
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
12–61
Reconsider Prob. 12–60. Using EES (or other)
software, investigate the effect of head loss on
mechanical efficiency of the pump. Let the head loss vary
from 0 to 15 m in increments of 1 m. Plot the results, and discuss them.
12–62 A 7-hp (shaft) pump is used to raise water to a 15-m
higher elevation. If the mechanical efficiency of the pump is 82
percent, determine the maximum volume flow rate of water.
12–63 Water flows at a rate of 0.035 m3/s in a horizontal pipe
whose diameter is reduced from 15 cm to 8 cm by a reducer. If
the pressure at the centerline is measured to be 470 kPa and
440 kPa before and after the reducer, respectively, determine
the head loss in the reducer. Answer: 0.79 m
generator during reverse flow. Preliminary analysis shows that
a water flow rate of 2 m3/s can be used in either direction, and
the head loss of the piping system is 4 m. The combined
pump-motor and turbine-generator efficiencies are expected to
be 75 percent each. Assuming the system operates for 10 h
each in the pump and turbine modes during a typical day, determine the potential revenue this pump-turbine system can
generate per year.
Reservoir
40 m
12–64 The water level in a tank is 20 m above the ground. A
hose is connected to the bottom of the tank, and the nozzle at
the end of the hose is pointed straight up. The tank is at sea
level, and the water surface is open to the atmosphere. In the
line leading from the tank to the nozzle is a pump, which increases the pressure of water. If the water jet rises to a height of
27 m from the ground, determine the minimum pressure rise
supplied by the pump to the water line.
27 m
20 m
PumpTurbine
Lake
FIGURE P12–66
12–67 Water flows at a rate of 20 L/s through a horizontal
pipe whose diameter is constant at 3 cm. The pressure drop
across a valve in the pipe is measured to be 2 kPa. Determine
the head loss of the valve, and the useful pumping power
needed to overcome the resulting pressure drop.
Answers: 0.204 m, 40 W
FIGURE P12–64
12–65 A hydraulic turbine has 85 m of head available at a
flow rate of 0.25 m3/s, and its overall turbine-generator efficiency is 78 percent. Determine the electric power output of
this turbine.
12–66 The demand for electric power is usually much higher
during the day than it is at night, and utility companies often
sell power at night at much lower prices to encourage consumers to use the available power generation capacity and to
avoid building new expensive power plants that will be used
only a short time during peak periods. Utilities are also willing
to purchase power produced during the day from private parties at a high price.
Suppose a utility company is selling electric power for
$0.03/kWh at night and is willing to pay $0.08/kWh for power
produced during the day. To take advantage of this opportunity, an entrepreneur is considering building a large reservoir
40 m above the lake level, pumping water from the lake to
the reservoir at night using cheap power, and letting the
water flow from the reservoir back to the lake during the day,
producing power as the pump-motor operates as a turbine-
Water
20 L/s
∆P = 2 kPa
FIGURE P12–67
12–68E The water level in a tank is 66 ft above the ground.
A hose is connected to the bottom of the tank, and the nozzle at
the end of the hose is pointed straight up. The tank cover is airtight, but the pressure over the water surface is unknown. Determine the minimum tank air pressure (gage) that will cause a
water stream from the nozzle to rise 90 ft from the ground.
12–69 A large tank is initially filled with water 2 m above
the center of a sharp-edged 10-cm-diameter orifice. The tank
water surface is open to the atmosphere, and the orifice drains
to the atmosphere. If the total head loss in the system is 0.3 m,
determine the initial discharge velocity of water from the tank.
557
CHAPTER 12
12–70 Water enters a hydraulic turbine through a 30-cmdiameter pipe at a rate of 0.6 m3/s and exits through a 25-cmdiameter pipe. The pressure drop in the turbine is measured by
a mercury manometer to be 1.2 m. For a combined turbinegenerator efficiency of 83 percent, determine the net electric
power output.
⋅
We
30 cm
Turbine
Generator
25 cm
∆P = 1.2 mHg
FIGURE P12–70
12–71 The velocity profile for turbulent flow in a circular pipe
is usually expressed as u(r) umax (1 r/R)1/n where n 7.
Determine the kinetic energy correction factor for this flow.
Answer: 1.06
12–72 An oil pump is drawing 35 kW of electric power while
pumping oil with r 860 kg/m3 at a rate of 0.1 m3/s. The inlet
and outlet diameters of the pipe are 8 cm and 12 cm, respectively. If the pressure rise of oil in the pump is measured to be
400 kPa and the motor efficiency is 90 percent, determine the
mechanical efficiency of the pump.
4m
FIGURE P12–74
Review Problems
12–75 A pressurized 2-m-diameter tank of water has a 10cm-diameter orifice at the bottom, where water discharges to
the atmosphere. The water level initially is 3 m above the outlet. The tank air pressure above the water level is maintained at
450 kPa absolute and the atmospheric pressure is 100 kPa. Neglecting frictional effects, determine (a) how long it will take
for half of the water in the tank to be discharged and (b) the
water level in the tank after 10 s.
12–76 Air flows through a pipe at a rate of 200 L/s. The pipe
consists of two sections of diameters 20 cm and 10 cm with a
smooth reducing section that connects them. The pressure difference between the two pipe sections is measured by a water
manometer. Neglecting frictional effects, determine the differential height of water between the two pipe sections. Take the
air density to be 1.20 kg/m3. Answer: 3.7 cm
Air
200 L/s
20 cm
10 cm
h
35 kW
12 cm
Pump
FIGURE P12–76
Motor
8 cm
Oil
0.1
∆P = 400 kPa
m3/s
FIGURE P12–72
12–73E A 73-percent efficient 12-hp pump is pumping water
from a lake to a nearby pool at a rate of 1.2 ft3/s through a
constant-diameter pipe. The free surface of the pool is 35 ft
above that of the lake. Determine the head loss of the piping
system, in ft, and the mechanical power used to overcome it.
12–74 A fireboat is to fight fires at coastal areas by drawing
seawater with a density of 1030 kg/m3 through a 20-cmdiameter pipe at a rate of 0.1 m3/s and discharging it through a
hose nozzle with an exit diameter of 5 cm. The total head loss
of the system is 3 m, and the position of the nozzle is 4 m
above sea level. For a pump efficiency of 70 percent, determine the required shaft power input to the pump and the water
discharge velocity. Answers: 200 kW, 50.9 m/s
12–77
Air at 100 kPa and 25C flows in a horizontal
duct of variable cross section. The water column
in the manometer that measures the difference between two
sections has a vertical displacement of 8 cm. If the velocity in
the first section is low and the friction is negligible, determine
the velocity at the second section. Also, if the manometer reading has a possible error of 2 mm, conduct an error analysis to
estimate the range of validity for the velocity found.
12–78 A very large tank contains air at 102 kPa at a
location where the atmospheric air is at 100 kPa and 20C.
Now a 2-cm-diameter tap is opened. Determine the maximum
flow rate of air through the hole. What would your response be
if air is discharged through a 2-m-long, 4-cm-diameter tube
with a 2-cm-diameter nozzle? Would you solve the problem the
same way if the pressure in the storage tank were 300 kPa?
12–79 Water is flowing through a venturi meter whose diameter is 7 cm at the entrance part and 4 cm at the throat. The
pressure is measured to be 430 kPa at the entrance and 120 kPa
at the throat. Neglecting frictional effects, determine the flow
rate of water. Answer: 0.538 m3/s
558
FUNDAMENTALS OF THERMAL-FLUID SCIENCES
100 kPa
20° C
Water
2 cm
Air
102 kPa
2m
2 cm
10 cm
4 cm
100 m
FIGURE P12–83
12–84
FIGURE P12–78
12–80E The water level in a tank is 80 ft above the ground.
A hose is connected to the bottom of the tank, and the nozzle at
the end of the hose is pointed straight up. The tank is at sea
level, and the water surface is open to the atmosphere. In the
line leading from the tank to the nozzle is a pump, which increases the water pressure by 10 psia. Determine the maximum
height to which the water stream could rise.
12–81 A wind tunnel draws atmospheric air at 20C and
101.3 kPa by a large fan located near the exit of the tunnel. If
the air velocity in the tunnel is 80 m/s, determine the pressure
in the tunnel.
20° C
101.3 kPa
Wind tunnel
80 m/s
FIGURE P12–81
12–82 Water flows at a rate of 0.025 m3/s in a horizontal pipe
whose diameter increases from 6 cm to 11 cm by an enlargement section. If the head loss across the enlargement section is
0.45 m, determine the pressure change.
12–83 A 2-m-high large tank is initially filled with water.
The tank water surface is open to the atmosphere, and a sharpedged 10-cm-diameter orifice at the bottom drains to the atmosphere through a horizontal 100-m-long pipe. If the total
head loss of the system is determined to be 1.5 m, determine
the initial velocity of the water from the tank.
Answer: 3.13 m/s
Reconsider Prob. 12–83. Using EES (or other)
software, investigate the effect of the tank height
on the initial discharge velocity of water from the completely
filled tank. Let the tank height vary from 2 to 15 m in increments of 1 m, and assume the head loss to remain constant.
Tabulate and plot the results.
12–85 Reconsider Prob. 12–83. In order to drain the tank
faster, a pump is installed near the tank exit. Determine the
pump head input necessary to establish an average water velocity of 6 m/s when the tank is full.
Design and Essay Problems
12–86 Your company is setting up an experiment that involves the measurement of airflow rate in a duct, and you are
to come up with proper instrumentation. Research the available
techniques and devices for airflow rate measurement, discuss
the advantages and disadvantages of each technique, and make
a recommendation.
12–87 Computer-aided designs, the use of better materials,
and better manufacturing techniques have resulted in a tremendous increase in the efficiency of pumps, turbines, and electric
motors. Contact several pump, turbine, and motor manufacturers and obtain information about the efficiency of their
products. In general, how does efficiency vary with rated
power of these devices?
12–88 Using a handheld bicycle pump to generate an air jet,
a soda can as the water reservoir, and a straw as the tube, design and build an atomizer. Study the effects of various parameters such as the tube length, the diameter of the exit hole,
and the pumping speed on performance.
12–89 Using a flexible drinking straw and a ruler, explain
how you would measure the water flow velocity in a river.